题目：现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况，请取出相应数据，并按照准确率升序输出。 示例： user_profile id device_id gender age university gpa active_days_within_30 question_cnt answer_cnt 1 2138 male 21 北京大学 3.4 7 2 12 2 3214 male 复旦大学 4 15 5 25 3 6543 female 20 北京大学 3.2 12 3 30 4 2315 female 23 浙江大学 3.6 5 1 2 5 5432 male 25 山东大学 3.8 20 15 70 6 2131 male 28 山东大学 3.3 15 7 13 7 4321 female 26 复旦大学 3.6 9 6 52 示例： question_practice_detail id device_id question_id result 1 2138 111 wrong 2 3214 112 wrong 3 3214 113 wrong 4 6543 111 right 5 2315 115 right 6 2315 116 right 7 2315 117 wrong 示例： question_detail question_id difficult_level 111 hard 112 medium 113 easy 115 easy 116 medium 117 easy 根据示例，你的查询应返回以下结果： difficult_level correct_rate easy 0.5000 medium 1.0000 示例1 输入 drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,’male’,21,’北京大学’,3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,’male’,null,’复旦大学’,4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,’female’,20,’北京大学’,3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,’female’,23,’浙江大学’,3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,’male’,25,’山东大学’,3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,’male’,28,’山东大学’,3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,’male’,28,’复旦大学’,3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,’wrong’,’2021-05-03′); INSERT INTO question_practice_detail VALUES(2,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(3,3214,113,’wrong’,’2021-06-15′); INSERT INTO question_practice_detail VALUES(4,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(5,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(6,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(7,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(8,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(9,3214,113,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(10,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(11,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(12,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(13,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(14,3214,112,’wrong’,’2021-08-16′); INSERT INTO question_practice_detail VALUES(15,3214,113,’wrong’,’2021-08-18′); INSERT INTO question_practice_detail VALUES(16,6543,111,’right’,’2021-08-13′); INSERT INTO question_detail VALUES(1,111,’hard’); INSERT INTO question_detail VALUES(2,112,’medium’); INSERT INTO question_detail VALUES(3,113,’easy’); INSERT INTO question_detail VALUES(4,115,’easy’); INSERT INTO question_detail VALUES(5,116,’medium’); INSERT INTO question_detail VALUES(6,117,’easy’); 输出 easy|0.5000 medium|1.0000 马上挑战 算法知识视频讲解 提交运行 算法知识视频讲解 添加笔记 求解答(0) 邀请回答 收藏(1250) 分享 提交结果有问题? 316个回答 120篇题解 添加回答 17 Mysql Qiuxj SELECT difficult_level, AVG(IF(result=’right’,1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university=’浙江大学’ GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-09-29 22:58:04 回复(4) 12 Mysql 老狗丶 select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate; 发表于 2021-08-28 16:11:28 回复(9) 4 智能路障 这题考察的是多表查询，只要把条件梳理到位就可迎刃而解 select difficult_level, count(if(result = ‘right’, 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = ‘浙江大学’ group by difficult_level order by correct_rate 发表于 2021-10-31 11:17:26 回复(2) 4 fire-light-guo select d.difficult_level, sum(if(result = ‘right’, 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = ‘浙江大学’ and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate 1. 做个三表关联 2. 按问题难度分组 3. sum（）嵌套if（）计算出正确的答题数量，再除以答题总数 发表于 2021-10-13 20:45:23 回复(2) 7 Baily1234 SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-10-15 16:22:28 回复(2) 3 茵桃小蚊子 没有上一题难… 现在习惯写代码前开始用加粗关键词和初步处理方法先捋一遍题目，然后就写得很顺手呢。如下： 现在运营想要了解浙江大学(where)的用户在不同难度题目(group by)下答题的正确率(正确数/题数)情况，请取出相应数据，并按照准确率升序(order by asc)输出。 SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC 发表于 2021-11-25 21:05:25 回复(0) 3 Mysql 其实是牛哥 select c.difficult_level, count(if(result=’right’,a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = ‘浙江大学’ )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate 发表于 2021-08-27 10:57:13 回复(0) 2 Mysql YonChun select difficult_level, count(if(result=’right’,1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university=’浙江大学’ group by difficult_level order by correct_rate 应该是最精简的了吧 发表于 2022-01-31 11:04:12 回复(0) 2 Mysql 牛客597002486号 select difficult_level, (count(if(result=’right’,1,null)) / count(a.question_id)) as correct_rate from question_practice_detail a join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id join question_detail c on a.question_id=c.question_id group by difficult_level order by correct_rate 发表于 2021-10-03 00:47:26 回复(4) 1 Mysql 369186344 select difficult_level, (count(case when t2.result = ‘right’ then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = ‘浙江大学’))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate; 发表于 2022-04-22 23:01:00 回复(0) 1 黑白配XY select difficult_level, count(if(result=’right’,result,null))/ count(result) from user_profile u join (select device_id,difficult_level,result from question_practice_detail q1 join question_detail q2 on q1.question_id = q2.question_id) t on u.device_id = t.device_id and university = ‘浙江大学’ group by difficult_level 发表于 2022-04-18 16:25:48 回复(0) 1 Mysql 牛客228167738号 SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC; 发表于 2022-04-16 23:45:52 回复(0) 1 Mysql 牛客398096850号 有没有哪位大神帮我看下我的none是哪步错了吗 SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = ‘right’ THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = ‘浙江大学’ )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate 发表于 2022-04-05 17:35:03 回复(0) 1 Mysql 柒小漠 select difficult_level, count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate from question_practice_detail join user_profile on question_practice_detail.device_id=user_profile.device_id join question_detail on question_practice_detail.question_id=question_detail.question_id where university = ‘浙江大学’ group by difficult_level order by rightrate 发表于 2022-03-29 14:55:59 回复(0) 1 Mysql Fayuan select t3.difficult_level,sum(case t2.result when ‘right’ then 1 else 0 end) / count(1) as ‘correct_rate’ from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id join question_detail t3 on t2.question_id=t3.question_id where t1.university=’浙江大学’ group by t3.difficult_level order by correct_rate 发表于 2022-03-22 15:07:27 回复(0) 1 XXCC111111 1、学位为浙江大学，为条件；2、不同难度水平的，就表示可能需要根据难度水平来分组，使用group by函数；3、答题正确率用sum(if(qpd.result = ‘right’, 1, 0))/count(*) 来计算 select difficult_level, sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate from question_practice_detail as qpd left join user_profile as up on qpd.device_id = up.device_id left join question_detail as qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by qd.difficult_level order by correct_rate; 发表于 2022-03-03 13:22:17 回复(0) 1 Mysql 柒鳴飞 select difficult_level, sum(if(result=’right’,1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by difficult_level order by correct_rate 发表于 2021-12-14 16:48:23 回复(0) 1 Mysql 牛客689071742号 SELECT d.difficult_level, sum( IF ( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate FROM question_practice_detail AS q INNER JOIN user_profile AS u ON u.device_id = q.device_id AND u.university = ‘浙江大学’ LEFT JOIN question_detail AS d ON q.question_id = d.question_id GROUP BY d.difficult_level ORDER BY correct_rate ASC 发表于 2021-11-20 11:22:56 回复(0) 1 Mysql 键盘盘盘盘 select difficult_level, sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate from question_practice_detail as qpd left join user_profile as u on u.device_id = qpd.device_id left join question_detail as qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by qd.difficult_level order by correct_rate 1注意写的时候如果都和中间的表有关,就得让中间的表left join其他的两个表 2注意count是统计行数,并不是计算,如果要计算的话,需要使用sum函数计算列里面的值!! 发表于 2021-11-08 14:50:46 回复(0) 1 Mysql 码农矿工小付 SELECT tmp.difficult_level, AVG(IF(tmp.result = ‘right’, 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = ‘浙江大学’) AS tmp GROUP BY tmp.difficult_level; 这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？ 发表于 2021-08-25 15:14:02 回复(7) 这道题你会答吗？花几分钟告诉大家答案吧！ 提交观点
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| id | device_id | gender | age | university | gpa | active_days_within_30 | question_cnt | answer_cnt |
| 1 | 2138 | male | 21 | 北京大学 | 3.4 | 7 | 2 | 12 |
| 2 | 3214 | male | 复旦大学 | 4 | 15 | 5 | 25 | |
| 3 | 6543 | female | 20 | 北京大学 | 3.2 | 12 | 3 | 30 |
| 4 | 2315 | female | 23 | 浙江大学 | 3.6 | 5 | 1 | 2 |
| 5 | 5432 | male | 25 | 山东大学 | 3.8 | 20 | 15 | 70 |
| 6 | 2131 | male | 28 | 山东大学 | 3.3 | 15 | 7 | 13 |
| 7 | 4321 | female | 26 | 复旦大学 | 3.6 | 9 | 6 | 52 |
| id | device_id | question_id | result |
| 1 | 2138 | 111 | wrong |
| 2 | 3214 | 112 | wrong |
| 3 | 3214 | 113 | wrong |
| 4 | 6543 | 111 | right |
| 5 | 2315 | 115 | right |
| 6 | 2315 | 116 | right |
| 7 | 2315 | 117 | wrong |
| question_id | difficult_level |
| 111 | hard |
| 112 | medium |
| 113 | easy |
| 115 | easy |
| 116 | medium |
| 117 | easy |
| difficult_level | correct_rate |
| easy | 0.5000 |
| medium | 1.0000 |
输入
drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,'male',null,'复旦大学',4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,'male',28,'山东大学',3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,'male',28,'复旦大学',3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,'wrong','2021-05-03'); INSERT INTO question_practice_detail VALUES(2,3214,112,'wrong','2021-05-09'); INSERT INTO question_practice_detail VALUES(3,3214,113,'wrong','2021-06-15'); INSERT INTO question_practice_detail VALUES(4,6543,111,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(5,2315,115,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(6,2315,116,'right','2021-08-14'); INSERT INTO question_practice_detail VALUES(7,2315,117,'wrong','2021-08-15'); INSERT INTO question_practice_detail VALUES(8,3214,112,'wrong','2021-05-09'); INSERT INTO question_practice_detail VALUES(9,3214,113,'wrong','2021-08-15'); INSERT INTO question_practice_detail VALUES(10,6543,111,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(11,2315,115,'right','2021-08-13'); INSERT INTO question_practice_detail VALUES(12,2315,116,'right','2021-08-14'); INSERT INTO question_practice_detail VALUES(13,2315,117,'wrong','2021-08-15'); INSERT INTO question_practice_detail VALUES(14,3214,112,'wrong','2021-08-16'); INSERT INTO question_practice_detail VALUES(15,3214,113,'wrong','2021-08-18'); INSERT INTO question_practice_detail VALUES(16,6543,111,'right','2021-08-13'); INSERT INTO question_detail VALUES(1,111,'hard'); INSERT INTO question_detail VALUES(2,112,'medium'); INSERT INTO question_detail VALUES(3,113,'easy'); INSERT INTO question_detail VALUES(4,115,'easy'); INSERT INTO question_detail VALUES(5,116,'medium'); INSERT INTO question_detail VALUES(6,117,'easy');
输出
easy|0.5000 medium|1.0000
SELECT difficult_level, AVG(IF(result='right',1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university='浙江大学' GROUP BY difficult_level ORDER BY correct_rate;
select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate;
select difficult_level, count(if(result = 'right', 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = '浙江大学' group by difficult_level order by correct_rate
select d.difficult_level, sum(if(result = 'right', 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = '浙江大学' and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate
SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate;
SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC
select c.difficult_level, count(if(result='right',a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = '浙江大学' )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate
select difficult_level, count(if(result='right',1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university='浙江大学' group by difficult_level order by correct_rate
应该是最精简的了吧
from
question_practice_detail a
join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id
join question_detail c on a.question_id=c.question_id
group by difficult_level
order by correct_rate
select difficult_level, (count(case when t2.result = 'right' then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = '浙江大学'))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate;
count(if(result=’right’,result,null))/ count(result)
from user_profile u
join
(select device_id,difficult_level,result
from question_practice_detail q1
join
question_detail q2
on q1.question_id = q2.question_id) t
on u.device_id = t.device_id and university = ‘浙江大学’
group by difficult_level
SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC;
SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = 'right' THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = '浙江大学' )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate
count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate
from question_practice_detail
join user_profile on question_practice_detail.device_id=user_profile.device_id
join question_detail on question_practice_detail.question_id=question_detail.question_id
where university = ‘浙江大学’
group by difficult_level
order by rightrate
from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id
join question_detail t3 on t2.question_id=t3.question_id
where t1.university=’浙江大学’
group by t3.difficult_level
order by correct_rate
sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate
from question_practice_detail as qpd
left join user_profile as up
on qpd.device_id = up.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate;
select difficult_level, sum(if(result='right',1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = '浙江大学' group by difficult_level order by correct_rate
d.difficult_level,
sum(
IF
( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate
FROM
question_practice_detail AS q
INNER JOIN user_profile AS u ON u.device_id = q.device_id
AND u.university = ‘浙江大学’
LEFT JOIN question_detail AS d ON q.question_id = d.question_id
GROUP BY
d.difficult_level
ORDER BY
correct_rate ASC
difficult_level,
sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate
from
question_practice_detail as qpd
left join
user_profile as u
on
u.device_id = qpd.device_id
left join
question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
SELECT tmp.difficult_level, AVG(IF(tmp.result = 'right', 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = '浙江大学') AS tmp GROUP BY tmp.difficult_level;
这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？
SELECT difficult_level, AVG(IF(result='right',1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university='浙江大学' GROUP BY difficult_level ORDER BY correct_rate;
select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate;
select difficult_level, count(if(result = 'right', 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = '浙江大学' group by difficult_level order by correct_rate
select d.difficult_level, sum(if(result = 'right', 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = '浙江大学' and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate
SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate;
SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC
select c.difficult_level, count(if(result='right',a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = '浙江大学' )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate
select difficult_level, count(if(result='right',1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university='浙江大学' group by difficult_level order by correct_rate
应该是最精简的了吧
from
question_practice_detail a
join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id
join question_detail c on a.question_id=c.question_id
group by difficult_level
order by correct_rate
select difficult_level, (count(case when t2.result = 'right' then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = '浙江大学'))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate;
count(if(result=’right’,result,null))/ count(result)
from user_profile u
join
(select device_id,difficult_level,result
from question_practice_detail q1
join
question_detail q2
on q1.question_id = q2.question_id) t
on u.device_id = t.device_id and university = ‘浙江大学’
group by difficult_level
SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC;
SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = 'right' THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = '浙江大学' )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate
count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate
from question_practice_detail
join user_profile on question_practice_detail.device_id=user_profile.device_id
join question_detail on question_practice_detail.question_id=question_detail.question_id
where university = ‘浙江大学’
group by difficult_level
order by rightrate
from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id
join question_detail t3 on t2.question_id=t3.question_id
where t1.university=’浙江大学’
group by t3.difficult_level
order by correct_rate
sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate
from question_practice_detail as qpd
left join user_profile as up
on qpd.device_id = up.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate;
select difficult_level, sum(if(result='right',1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = '浙江大学' group by difficult_level order by correct_rate
d.difficult_level,
sum(
IF
( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate
FROM
question_practice_detail AS q
INNER JOIN user_profile AS u ON u.device_id = q.device_id
AND u.university = ‘浙江大学’
LEFT JOIN question_detail AS d ON q.question_id = d.question_id
GROUP BY
d.difficult_level
ORDER BY
correct_rate ASC
difficult_level,
sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate
from
question_practice_detail as qpd
left join
user_profile as u
on
u.device_id = qpd.device_id
left join
question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
SELECT tmp.difficult_level, AVG(IF(tmp.result = 'right', 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = '浙江大学') AS tmp GROUP BY tmp.difficult_level;
这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？
以上就是关于问题题目：现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况，请取出相应数据，并按照准确率升序输出。
示例： user_profile id device_id gender age university gpa active_days_within_30
question_cnt
answer_cnt 1 2138 male 21 北京大学 3.4 7 2 12 2 3214 male 复旦大学 4 15 5 25 3 6543 female 20 北京大学 3.2 12 3 30 4 2315 female 23 浙江大学 3.6 5 1 2 5 5432 male 25 山东大学 3.8 20 15 70 6 2131 male 28 山东大学 3.3 15 7 13 7 4321 female 26 复旦大学 3.6 9 6 52
示例： question_practice_detail id device_id question_id result 1 2138 111 wrong 2 3214 112 wrong
3 3214 113 wrong
4 6543 111 right 5 2315 115 right
6 2315 116 right
7 2315 117 wrong
示例： question_detail question_id difficult_level 111 hard 112 medium
113 easy
115 easy
116 medium
117 easy
根据示例，你的查询应返回以下结果： difficult_level correct_rate easy
0.5000 medium
1.0000
示例1 输入 drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,’male’,21,’北京大学’,3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,’male’,null,’复旦大学’,4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,’female’,20,’北京大学’,3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,’female’,23,’浙江大学’,3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,’male’,25,’山东大学’,3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,’male’,28,’山东大学’,3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,’male’,28,’复旦大学’,3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,’wrong’,’2021-05-03′); INSERT INTO question_practice_detail VALUES(2,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(3,3214,113,’wrong’,’2021-06-15′); INSERT INTO question_practice_detail VALUES(4,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(5,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(6,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(7,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(8,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(9,3214,113,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(10,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(11,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(12,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(13,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(14,3214,112,’wrong’,’2021-08-16′); INSERT INTO question_practice_detail VALUES(15,3214,113,’wrong’,’2021-08-18′); INSERT INTO question_practice_detail VALUES(16,6543,111,’right’,’2021-08-13′); INSERT INTO question_detail VALUES(1,111,’hard’); INSERT INTO question_detail VALUES(2,112,’medium’); INSERT INTO question_detail VALUES(3,113,’easy’); INSERT INTO question_detail VALUES(4,115,’easy’); INSERT INTO question_detail VALUES(5,116,’medium’); INSERT INTO question_detail VALUES(6,117,’easy’); 输出 easy|0.5000 medium|1.0000 马上挑战 算法知识视频讲解
提交运行 算法知识视频讲解 添加笔记 求解答(0) 邀请回答 收藏(1250) 分享 提交结果有问题? 316个回答 120篇题解 添加回答 17 Mysql Qiuxj SELECT difficult_level, AVG(IF(result=’right’,1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university=’浙江大学’ GROUP BY difficult_level ORDER BY correct_rate;
发表于 2021-09-29 22:58:04 回复(4) 12 Mysql 老狗丶 select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate;
发表于 2021-08-28 16:11:28 回复(9) 4 智能路障 这题考察的是多表查询，只要把条件梳理到位就可迎刃而解 select difficult_level, count(if(result = ‘right’, 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-10-31 11:17:26 回复(2) 4 fire-light-guo select d.difficult_level, sum(if(result = ‘right’, 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = ‘浙江大学’ and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate 1. 做个三表关联 2. 按问题难度分组 3. sum（）嵌套if（）计算出正确的答题数量，再除以答题总数
发表于 2021-10-13 20:45:23 回复(2) 7 Baily1234 SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-10-15 16:22:28 回复(2) 3 茵桃小蚊子 没有上一题难… 现在习惯写代码前开始用加粗关键词和初步处理方法先捋一遍题目，然后就写得很顺手呢。如下： 现在运营想要了解浙江大学(where)的用户在不同难度题目(group by)下答题的正确率(正确数/题数)情况，请取出相应数据，并按照准确率升序(order by asc)输出。
SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC
发表于 2021-11-25 21:05:25 回复(0) 3 Mysql 其实是牛哥 select c.difficult_level, count(if(result=’right’,a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = ‘浙江大学’ )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate
发表于 2021-08-27 10:57:13 回复(0) 2 Mysql YonChun select difficult_level, count(if(result=’right’,1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university=’浙江大学’ group by difficult_level order by correct_rate 应该是最精简的了吧 发表于 2022-01-31 11:04:12 回复(0) 2 Mysql 牛客597002486号 select difficult_level, (count(if(result=’right’,1,null)) / count(a.question_id)) as correct_rate
from
question_practice_detail a
join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id
join question_detail c on a.question_id=c.question_id
group by difficult_level
order by correct_rate
发表于 2021-10-03 00:47:26 回复(4) 1 Mysql 369186344 select difficult_level, (count(case when t2.result = ‘right’ then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = ‘浙江大学’))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate;
发表于 2022-04-22 23:01:00 回复(0) 1 黑白配XY select difficult_level,
count(if(result=’right’,result,null))/ count(result)
from user_profile u
join
(select device_id,difficult_level,result
from question_practice_detail q1
join
question_detail q2
on q1.question_id = q2.question_id) t
on u.device_id = t.device_id and university = ‘浙江大学’
group by difficult_level
发表于 2022-04-18 16:25:48 回复(0) 1 Mysql 牛客228167738号 SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC; 发表于 2022-04-16 23:45:52 回复(0) 1 Mysql 牛客398096850号 有没有哪位大神帮我看下我的none是哪步错了吗
SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = ‘right’ THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = ‘浙江大学’ )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate
发表于 2022-04-05 17:35:03 回复(0) 1 Mysql 柒小漠 select difficult_level,
count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate
from question_practice_detail
join user_profile on question_practice_detail.device_id=user_profile.device_id
join question_detail on question_practice_detail.question_id=question_detail.question_id
where university = ‘浙江大学’
group by difficult_level
order by rightrate
发表于 2022-03-29 14:55:59 回复(0) 1 Mysql Fayuan select t3.difficult_level,sum(case t2.result when ‘right’ then 1 else 0 end) / count(1) as ‘correct_rate’
from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id
join question_detail t3 on t2.question_id=t3.question_id
where t1.university=’浙江大学’
group by t3.difficult_level
order by correct_rate
发表于 2022-03-22 15:07:27 回复(0) 1 XXCC111111 1、学位为浙江大学，为条件；2、不同难度水平的，就表示可能需要根据难度水平来分组，使用group by函数；3、答题正确率用sum(if(qpd.result = ‘right’, 1, 0))/count(*) 来计算
select difficult_level,
sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate
from question_practice_detail as qpd
left join user_profile as up
on qpd.device_id = up.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate;
发表于 2022-03-03 13:22:17 回复(0) 1 Mysql 柒鳴飞 select difficult_level, sum(if(result=’right’,1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-12-14 16:48:23 回复(0) 1 Mysql 牛客689071742号 SELECT
d.difficult_level,
sum(
IF
( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate
FROM
question_practice_detail AS q
INNER JOIN user_profile AS u ON u.device_id = q.device_id
AND u.university = ‘浙江大学’
LEFT JOIN question_detail AS d ON q.question_id = d.question_id
GROUP BY
d.difficult_level
ORDER BY
correct_rate ASC
发表于 2021-11-20 11:22:56 回复(0) 1 Mysql 键盘盘盘盘 select
difficult_level,
sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate
from
question_practice_detail as qpd
left join
user_profile as u
on
u.device_id = qpd.device_id
left join
question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate
1注意写的时候如果都和中间的表有关,就得让中间的表left join其他的两个表 2注意count是统计行数,并不是计算,如果要计算的话,需要使用sum函数计算列里面的值!! 发表于 2021-11-08 14:50:46 回复(0) 1 Mysql 码农矿工小付 SELECT tmp.difficult_level, AVG(IF(tmp.result = ‘right’, 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = ‘浙江大学’) AS tmp GROUP BY tmp.difficult_level; 这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？ 发表于 2021-08-25 15:14:02 回复(7) 这道题你会答吗？花几分钟告诉大家答案吧！ 提交观点的答案
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qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 题目：现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况，请取出相应数据，并按照准确率升序输出。
示例： user_profile id device_id gender age university gpa active_days_within_30
question_cnt
answer_cnt 1 2138 male 21 北京大学 3.4 7 2 12 2 3214 male 复旦大学 4 15 5 25 3 6543 female 20 北京大学 3.2 12 3 30 4 2315 female 23 浙江大学 3.6 5 1 2 5 5432 male 25 山东大学 3.8 20 15 70 6 2131 male 28 山东大学 3.3 15 7 13 7 4321 female 26 复旦大学 3.6 9 6 52
示例： question_practice_detail id device_id question_id result 1 2138 111 wrong 2 3214 112 wrong
3 3214 113 wrong
4 6543 111 right 5 2315 115 right
6 2315 116 right
7 2315 117 wrong
示例： question_detail question_id difficult_level 111 hard 112 medium
113 easy
115 easy
116 medium
117 easy
根据示例，你的查询应返回以下结果： difficult_level correct_rate easy
0.5000 medium
1.0000
示例1 输入 drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,’male’,21,’北京大学’,3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,’male’,null,’复旦大学’,4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,’female’,20,’北京大学’,3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,’female’,23,’浙江大学’,3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,’male’,25,’山东大学’,3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,’male’,28,’山东大学’,3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,’male’,28,’复旦大学’,3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,’wrong’,’2021-05-03′); INSERT INTO question_practice_detail VALUES(2,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(3,3214,113,’wrong’,’2021-06-15′); INSERT INTO question_practice_detail VALUES(4,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(5,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(6,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(7,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(8,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(9,3214,113,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(10,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(11,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(12,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(13,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(14,3214,112,’wrong’,’2021-08-16′); INSERT INTO question_practice_detail VALUES(15,3214,113,’wrong’,’2021-08-18′); INSERT INTO question_practice_detail VALUES(16,6543,111,’right’,’2021-08-13′); INSERT INTO question_detail VALUES(1,111,’hard’); INSERT INTO question_detail VALUES(2,112,’medium’); INSERT INTO question_detail VALUES(3,113,’easy’); INSERT INTO question_detail VALUES(4,115,’easy’); INSERT INTO question_detail VALUES(5,116,’medium’); INSERT INTO question_detail VALUES(6,117,’easy’); 输出 easy|0.5000 medium|1.0000 马上挑战 算法知识视频讲解
提交运行 算法知识视频讲解 添加笔记 求解答(0) 邀请回答 收藏(1250) 分享 提交结果有问题? 316个回答 120篇题解 添加回答 17 Mysql Qiuxj SELECT difficult_level, AVG(IF(result=’right’,1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university=’浙江大学’ GROUP BY difficult_level ORDER BY correct_rate;
发表于 2021-09-29 22:58:04 回复(4) 12 Mysql 老狗丶 select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate;
发表于 2021-08-28 16:11:28 回复(9) 4 智能路障 这题考察的是多表查询，只要把条件梳理到位就可迎刃而解 select difficult_level, count(if(result = ‘right’, 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-10-31 11:17:26 回复(2) 4 fire-light-guo select d.difficult_level, sum(if(result = ‘right’, 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = ‘浙江大学’ and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate 1. 做个三表关联 2. 按问题难度分组 3. sum（）嵌套if（）计算出正确的答题数量，再除以答题总数
发表于 2021-10-13 20:45:23 回复(2) 7 Baily1234 SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-10-15 16:22:28 回复(2) 3 茵桃小蚊子 没有上一题难… 现在习惯写代码前开始用加粗关键词和初步处理方法先捋一遍题目，然后就写得很顺手呢。如下： 现在运营想要了解浙江大学(where)的用户在不同难度题目(group by)下答题的正确率(正确数/题数)情况，请取出相应数据，并按照准确率升序(order by asc)输出。
SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC
发表于 2021-11-25 21:05:25 回复(0) 3 Mysql 其实是牛哥 select c.difficult_level, count(if(result=’right’,a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = ‘浙江大学’ )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate
发表于 2021-08-27 10:57:13 回复(0) 2 Mysql YonChun select difficult_level, count(if(result=’right’,1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university=’浙江大学’ group by difficult_level order by correct_rate 应该是最精简的了吧 发表于 2022-01-31 11:04:12 回复(0) 2 Mysql 牛客597002486号 select difficult_level, (count(if(result=’right’,1,null)) / count(a.question_id)) as correct_rate
from
question_practice_detail a
join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id
join question_detail c on a.question_id=c.question_id
group by difficult_level
order by correct_rate
发表于 2021-10-03 00:47:26 回复(4) 1 Mysql 369186344 select difficult_level, (count(case when t2.result = ‘right’ then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = ‘浙江大学’))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate;
发表于 2022-04-22 23:01:00 回复(0) 1 黑白配XY select difficult_level,
count(if(result=’right’,result,null))/ count(result)
from user_profile u
join
(select device_id,difficult_level,result
from question_practice_detail q1
join
question_detail q2
on q1.question_id = q2.question_id) t
on u.device_id = t.device_id and university = ‘浙江大学’
group by difficult_level
发表于 2022-04-18 16:25:48 回复(0) 1 Mysql 牛客228167738号 SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC; 发表于 2022-04-16 23:45:52 回复(0) 1 Mysql 牛客398096850号 有没有哪位大神帮我看下我的none是哪步错了吗
SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = ‘right’ THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = ‘浙江大学’ )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate
发表于 2022-04-05 17:35:03 回复(0) 1 Mysql 柒小漠 select difficult_level,
count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate
from question_practice_detail
join user_profile on question_practice_detail.device_id=user_profile.device_id
join question_detail on question_practice_detail.question_id=question_detail.question_id
where university = ‘浙江大学’
group by difficult_level
order by rightrate
发表于 2022-03-29 14:55:59 回复(0) 1 Mysql Fayuan select t3.difficult_level,sum(case t2.result when ‘right’ then 1 else 0 end) / count(1) as ‘correct_rate’
from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id
join question_detail t3 on t2.question_id=t3.question_id
where t1.university=’浙江大学’
group by t3.difficult_level
order by correct_rate
发表于 2022-03-22 15:07:27 回复(0) 1 XXCC111111 1、学位为浙江大学，为条件；2、不同难度水平的，就表示可能需要根据难度水平来分组，使用group by函数；3、答题正确率用sum(if(qpd.result = ‘right’, 1, 0))/count(*) 来计算
select difficult_level,
sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate
from question_practice_detail as qpd
left join user_profile as up
on qpd.device_id = up.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate;
发表于 2022-03-03 13:22:17 回复(0) 1 Mysql 柒鳴飞 select difficult_level, sum(if(result=’right’,1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-12-14 16:48:23 回复(0) 1 Mysql 牛客689071742号 SELECT
d.difficult_level,
sum(
IF
( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate
FROM
question_practice_detail AS q
INNER JOIN user_profile AS u ON u.device_id = q.device_id
AND u.university = ‘浙江大学’
LEFT JOIN question_detail AS d ON q.question_id = d.question_id
GROUP BY
d.difficult_level
ORDER BY
correct_rate ASC
发表于 2021-11-20 11:22:56 回复(0) 1 Mysql 键盘盘盘盘 select
difficult_level,
sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate
from
question_practice_detail as qpd
left join
user_profile as u
on
u.device_id = qpd.device_id
left join
question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate
1注意写的时候如果都和中间的表有关,就得让中间的表left join其他的两个表 2注意count是统计行数,并不是计算,如果要计算的话,需要使用sum函数计算列里面的值!! 发表于 2021-11-08 14:50:46 回复(0) 1 Mysql 码农矿工小付 SELECT tmp.difficult_level, AVG(IF(tmp.result = ‘right’, 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = ‘浙江大学’) AS tmp GROUP BY tmp.difficult_level; 这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？ 发表于 2021-08-25 15:14:02 回复(7) 这道题你会答吗？花几分钟告诉大家答案吧！ 提交观点
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qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 题目：现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况，请取出相应数据，并按照准确率升序输出。
示例： user_profile id device_id gender age university gpa active_days_within_30
question_cnt
answer_cnt 1 2138 male 21 北京大学 3.4 7 2 12 2 3214 male 复旦大学 4 15 5 25 3 6543 female 20 北京大学 3.2 12 3 30 4 2315 female 23 浙江大学 3.6 5 1 2 5 5432 male 25 山东大学 3.8 20 15 70 6 2131 male 28 山东大学 3.3 15 7 13 7 4321 female 26 复旦大学 3.6 9 6 52
示例： question_practice_detail id device_id question_id result 1 2138 111 wrong 2 3214 112 wrong
3 3214 113 wrong
4 6543 111 right 5 2315 115 right
6 2315 116 right
7 2315 117 wrong
示例： question_detail question_id difficult_level 111 hard 112 medium
113 easy
115 easy
116 medium
117 easy
根据示例，你的查询应返回以下结果： difficult_level correct_rate easy
0.5000 medium
1.0000
示例1 输入 drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,’male’,21,’北京大学’,3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,’male’,null,’复旦大学’,4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,’female’,20,’北京大学’,3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,’female’,23,’浙江大学’,3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,’male’,25,’山东大学’,3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,’male’,28,’山东大学’,3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,’male’,28,’复旦大学’,3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,’wrong’,’2021-05-03′); INSERT INTO question_practice_detail VALUES(2,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(3,3214,113,’wrong’,’2021-06-15′); INSERT INTO question_practice_detail VALUES(4,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(5,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(6,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(7,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(8,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(9,3214,113,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(10,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(11,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(12,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(13,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(14,3214,112,’wrong’,’2021-08-16′); INSERT INTO question_practice_detail VALUES(15,3214,113,’wrong’,’2021-08-18′); INSERT INTO question_practice_detail VALUES(16,6543,111,’right’,’2021-08-13′); INSERT INTO question_detail VALUES(1,111,’hard’); INSERT INTO question_detail VALUES(2,112,’medium’); INSERT INTO question_detail VALUES(3,113,’easy’); INSERT INTO question_detail VALUES(4,115,’easy’); INSERT INTO question_detail VALUES(5,116,’medium’); INSERT INTO question_detail VALUES(6,117,’easy’); 输出 easy|0.5000 medium|1.0000 马上挑战 算法知识视频讲解
提交运行 算法知识视频讲解 添加笔记 求解答(0) 邀请回答 收藏(1250) 分享 提交结果有问题? 316个回答 120篇题解 添加回答 17 Mysql Qiuxj SELECT difficult_level, AVG(IF(result=’right’,1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university=’浙江大学’ GROUP BY difficult_level ORDER BY correct_rate;
发表于 2021-09-29 22:58:04 回复(4) 12 Mysql 老狗丶 select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate;
发表于 2021-08-28 16:11:28 回复(9) 4 智能路障 这题考察的是多表查询，只要把条件梳理到位就可迎刃而解 select difficult_level, count(if(result = ‘right’, 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-10-31 11:17:26 回复(2) 4 fire-light-guo select d.difficult_level, sum(if(result = ‘right’, 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = ‘浙江大学’ and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate 1. 做个三表关联 2. 按问题难度分组 3. sum（）嵌套if（）计算出正确的答题数量，再除以答题总数
发表于 2021-10-13 20:45:23 回复(2) 7 Baily1234 SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-10-15 16:22:28 回复(2) 3 茵桃小蚊子 没有上一题难… 现在习惯写代码前开始用加粗关键词和初步处理方法先捋一遍题目，然后就写得很顺手呢。如下： 现在运营想要了解浙江大学(where)的用户在不同难度题目(group by)下答题的正确率(正确数/题数)情况，请取出相应数据，并按照准确率升序(order by asc)输出。
SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC
发表于 2021-11-25 21:05:25 回复(0) 3 Mysql 其实是牛哥 select c.difficult_level, count(if(result=’right’,a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = ‘浙江大学’ )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate
发表于 2021-08-27 10:57:13 回复(0) 2 Mysql YonChun select difficult_level, count(if(result=’right’,1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university=’浙江大学’ group by difficult_level order by correct_rate 应该是最精简的了吧 发表于 2022-01-31 11:04:12 回复(0) 2 Mysql 牛客597002486号 select difficult_level, (count(if(result=’right’,1,null)) / count(a.question_id)) as correct_rate
from
question_practice_detail a
join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id
join question_detail c on a.question_id=c.question_id
group by difficult_level
order by correct_rate
发表于 2021-10-03 00:47:26 回复(4) 1 Mysql 369186344 select difficult_level, (count(case when t2.result = ‘right’ then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = ‘浙江大学’))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate;
发表于 2022-04-22 23:01:00 回复(0) 1 黑白配XY select difficult_level,
count(if(result=’right’,result,null))/ count(result)
from user_profile u
join
(select device_id,difficult_level,result
from question_practice_detail q1
join
question_detail q2
on q1.question_id = q2.question_id) t
on u.device_id = t.device_id and university = ‘浙江大学’
group by difficult_level
发表于 2022-04-18 16:25:48 回复(0) 1 Mysql 牛客228167738号 SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC; 发表于 2022-04-16 23:45:52 回复(0) 1 Mysql 牛客398096850号 有没有哪位大神帮我看下我的none是哪步错了吗
SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = ‘right’ THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = ‘浙江大学’ )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate
发表于 2022-04-05 17:35:03 回复(0) 1 Mysql 柒小漠 select difficult_level,
count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate
from question_practice_detail
join user_profile on question_practice_detail.device_id=user_profile.device_id
join question_detail on question_practice_detail.question_id=question_detail.question_id
where university = ‘浙江大学’
group by difficult_level
order by rightrate
发表于 2022-03-29 14:55:59 回复(0) 1 Mysql Fayuan select t3.difficult_level,sum(case t2.result when ‘right’ then 1 else 0 end) / count(1) as ‘correct_rate’
from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id
join question_detail t3 on t2.question_id=t3.question_id
where t1.university=’浙江大学’
group by t3.difficult_level
order by correct_rate
发表于 2022-03-22 15:07:27 回复(0) 1 XXCC111111 1、学位为浙江大学，为条件；2、不同难度水平的，就表示可能需要根据难度水平来分组，使用group by函数；3、答题正确率用sum(if(qpd.result = ‘right’, 1, 0))/count(*) 来计算
select difficult_level,
sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate
from question_practice_detail as qpd
left join user_profile as up
on qpd.device_id = up.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate;
发表于 2022-03-03 13:22:17 回复(0) 1 Mysql 柒鳴飞 select difficult_level, sum(if(result=’right’,1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-12-14 16:48:23 回复(0) 1 Mysql 牛客689071742号 SELECT
d.difficult_level,
sum(
IF
( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate
FROM
question_practice_detail AS q
INNER JOIN user_profile AS u ON u.device_id = q.device_id
AND u.university = ‘浙江大学’
LEFT JOIN question_detail AS d ON q.question_id = d.question_id
GROUP BY
d.difficult_level
ORDER BY
correct_rate ASC
发表于 2021-11-20 11:22:56 回复(0) 1 Mysql 键盘盘盘盘 select
difficult_level,
sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate
from
question_practice_detail as qpd
left join
user_profile as u
on
u.device_id = qpd.device_id
left join
question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate
1注意写的时候如果都和中间的表有关,就得让中间的表left join其他的两个表 2注意count是统计行数,并不是计算,如果要计算的话,需要使用sum函数计算列里面的值!! 发表于 2021-11-08 14:50:46 回复(0) 1 Mysql 码农矿工小付 SELECT tmp.difficult_level, AVG(IF(tmp.result = ‘right’, 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = ‘浙江大学’) AS tmp GROUP BY tmp.difficult_level; 这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？ 发表于 2021-08-25 15:14:02 回复(7) 这道题你会答吗？花几分钟告诉大家答案吧！ 提交观点
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qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 题目：现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况，请取出相应数据，并按照准确率升序输出。
示例： user_profile id device_id gender age university gpa active_days_within_30
question_cnt
answer_cnt 1 2138 male 21 北京大学 3.4 7 2 12 2 3214 male 复旦大学 4 15 5 25 3 6543 female 20 北京大学 3.2 12 3 30 4 2315 female 23 浙江大学 3.6 5 1 2 5 5432 male 25 山东大学 3.8 20 15 70 6 2131 male 28 山东大学 3.3 15 7 13 7 4321 female 26 复旦大学 3.6 9 6 52
示例： question_practice_detail id device_id question_id result 1 2138 111 wrong 2 3214 112 wrong
3 3214 113 wrong
4 6543 111 right 5 2315 115 right
6 2315 116 right
7 2315 117 wrong
示例： question_detail question_id difficult_level 111 hard 112 medium
113 easy
115 easy
116 medium
117 easy
根据示例，你的查询应返回以下结果： difficult_level correct_rate easy
0.5000 medium
1.0000
示例1 输入 drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,’male’,21,’北京大学’,3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,’male’,null,’复旦大学’,4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,’female’,20,’北京大学’,3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,’female’,23,’浙江大学’,3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,’male’,25,’山东大学’,3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,’male’,28,’山东大学’,3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,’male’,28,’复旦大学’,3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,’wrong’,’2021-05-03′); INSERT INTO question_practice_detail VALUES(2,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(3,3214,113,’wrong’,’2021-06-15′); INSERT INTO question_practice_detail VALUES(4,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(5,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(6,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(7,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(8,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(9,3214,113,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(10,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(11,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(12,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(13,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(14,3214,112,’wrong’,’2021-08-16′); INSERT INTO question_practice_detail VALUES(15,3214,113,’wrong’,’2021-08-18′); INSERT INTO question_practice_detail VALUES(16,6543,111,’right’,’2021-08-13′); INSERT INTO question_detail VALUES(1,111,’hard’); INSERT INTO question_detail VALUES(2,112,’medium’); INSERT INTO question_detail VALUES(3,113,’easy’); INSERT INTO question_detail VALUES(4,115,’easy’); INSERT INTO question_detail VALUES(5,116,’medium’); INSERT INTO question_detail VALUES(6,117,’easy’); 输出 easy|0.5000 medium|1.0000 马上挑战 算法知识视频讲解
提交运行 算法知识视频讲解 添加笔记 求解答(0) 邀请回答 收藏(1250) 分享 提交结果有问题? 316个回答 120篇题解 添加回答 17 Mysql Qiuxj SELECT difficult_level, AVG(IF(result=’right’,1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university=’浙江大学’ GROUP BY difficult_level ORDER BY correct_rate;
发表于 2021-09-29 22:58:04 回复(4) 12 Mysql 老狗丶 select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate;
发表于 2021-08-28 16:11:28 回复(9) 4 智能路障 这题考察的是多表查询，只要把条件梳理到位就可迎刃而解 select difficult_level, count(if(result = ‘right’, 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-10-31 11:17:26 回复(2) 4 fire-light-guo select d.difficult_level, sum(if(result = ‘right’, 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = ‘浙江大学’ and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate 1. 做个三表关联 2. 按问题难度分组 3. sum（）嵌套if（）计算出正确的答题数量，再除以答题总数
发表于 2021-10-13 20:45:23 回复(2) 7 Baily1234 SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-10-15 16:22:28 回复(2) 3 茵桃小蚊子 没有上一题难… 现在习惯写代码前开始用加粗关键词和初步处理方法先捋一遍题目，然后就写得很顺手呢。如下： 现在运营想要了解浙江大学(where)的用户在不同难度题目(group by)下答题的正确率(正确数/题数)情况，请取出相应数据，并按照准确率升序(order by asc)输出。
SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC
发表于 2021-11-25 21:05:25 回复(0) 3 Mysql 其实是牛哥 select c.difficult_level, count(if(result=’right’,a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = ‘浙江大学’ )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate
发表于 2021-08-27 10:57:13 回复(0) 2 Mysql YonChun select difficult_level, count(if(result=’right’,1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university=’浙江大学’ group by difficult_level order by correct_rate 应该是最精简的了吧 发表于 2022-01-31 11:04:12 回复(0) 2 Mysql 牛客597002486号 select difficult_level, (count(if(result=’right’,1,null)) / count(a.question_id)) as correct_rate
from
question_practice_detail a
join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id
join question_detail c on a.question_id=c.question_id
group by difficult_level
order by correct_rate
发表于 2021-10-03 00:47:26 回复(4) 1 Mysql 369186344 select difficult_level, (count(case when t2.result = ‘right’ then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = ‘浙江大学’))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate;
发表于 2022-04-22 23:01:00 回复(0) 1 黑白配XY select difficult_level,
count(if(result=’right’,result,null))/ count(result)
from user_profile u
join
(select device_id,difficult_level,result
from question_practice_detail q1
join
question_detail q2
on q1.question_id = q2.question_id) t
on u.device_id = t.device_id and university = ‘浙江大学’
group by difficult_level
发表于 2022-04-18 16:25:48 回复(0) 1 Mysql 牛客228167738号 SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC; 发表于 2022-04-16 23:45:52 回复(0) 1 Mysql 牛客398096850号 有没有哪位大神帮我看下我的none是哪步错了吗
SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = ‘right’ THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = ‘浙江大学’ )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate
发表于 2022-04-05 17:35:03 回复(0) 1 Mysql 柒小漠 select difficult_level,
count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate
from question_practice_detail
join user_profile on question_practice_detail.device_id=user_profile.device_id
join question_detail on question_practice_detail.question_id=question_detail.question_id
where university = ‘浙江大学’
group by difficult_level
order by rightrate
发表于 2022-03-29 14:55:59 回复(0) 1 Mysql Fayuan select t3.difficult_level,sum(case t2.result when ‘right’ then 1 else 0 end) / count(1) as ‘correct_rate’
from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id
join question_detail t3 on t2.question_id=t3.question_id
where t1.university=’浙江大学’
group by t3.difficult_level
order by correct_rate
发表于 2022-03-22 15:07:27 回复(0) 1 XXCC111111 1、学位为浙江大学，为条件；2、不同难度水平的，就表示可能需要根据难度水平来分组，使用group by函数；3、答题正确率用sum(if(qpd.result = ‘right’, 1, 0))/count(*) 来计算
select difficult_level,
sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate
from question_practice_detail as qpd
left join user_profile as up
on qpd.device_id = up.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate;
发表于 2022-03-03 13:22:17 回复(0) 1 Mysql 柒鳴飞 select difficult_level, sum(if(result=’right’,1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by difficult_level order by correct_rate
发表于 2021-12-14 16:48:23 回复(0) 1 Mysql 牛客689071742号 SELECT
d.difficult_level,
sum(
IF
( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate
FROM
question_practice_detail AS q
INNER JOIN user_profile AS u ON u.device_id = q.device_id
AND u.university = ‘浙江大学’
LEFT JOIN question_detail AS d ON q.question_id = d.question_id
GROUP BY
d.difficult_level
ORDER BY
correct_rate ASC
发表于 2021-11-20 11:22:56 回复(0) 1 Mysql 键盘盘盘盘 select
difficult_level,
sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate
from
question_practice_detail as qpd
left join
user_profile as u
on
u.device_id = qpd.device_id
left join
question_detail as qd
on qpd.question_id = qd.question_id
where university = ‘浙江大学’
group by qd.difficult_level
order by correct_rate
1注意写的时候如果都和中间的表有关,就得让中间的表left join其他的两个表 2注意count是统计行数,并不是计算,如果要计算的话,需要使用sum函数计算列里面的值!! 发表于 2021-11-08 14:50:46 回复(0) 1 Mysql 码农矿工小付 SELECT tmp.difficult_level, AVG(IF(tmp.result = ‘right’, 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = ‘浙江大学’) AS tmp GROUP BY tmp.difficult_level; 这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？ 发表于 2021-08-25 15:14:02 回复(7) 这道题你会答吗？花几分钟告诉大家答案吧！ 提交观点
qklbishe.com区块链毕设代做网专注|以太坊fabric-计算机|java|毕业设计|代做平台-javagopython毕设 » 题目：现在运营想要了解浙江大学的用户在不同难度题目下答题的正确率情况，请取出相应数据，并按照准确率升序输出。 示例： user_profile id device_id gender age university gpa active_days_within_30 question_cnt answer_cnt 1 2138 male 21 北京大学 3.4 7 2 12 2 3214 male 复旦大学 4 15 5 25 3 6543 female 20 北京大学 3.2 12 3 30 4 2315 female 23 浙江大学 3.6 5 1 2 5 5432 male 25 山东大学 3.8 20 15 70 6 2131 male 28 山东大学 3.3 15 7 13 7 4321 female 26 复旦大学 3.6 9 6 52 示例： question_practice_detail id device_id question_id result 1 2138 111 wrong 2 3214 112 wrong 3 3214 113 wrong 4 6543 111 right 5 2315 115 right 6 2315 116 right 7 2315 117 wrong 示例： question_detail question_id difficult_level 111 hard 112 medium 113 easy 115 easy 116 medium 117 easy 根据示例，你的查询应返回以下结果： difficult_level correct_rate easy 0.5000 medium 1.0000 示例1 输入 drop table if exists `user_profile`; drop table if exists `question_practice_detail`; drop table if exists `question_detail`; CREATE TABLE `user_profile` ( `id` int NOT NULL, `device_id` int NOT NULL, `gender` varchar(14) NOT NULL, `age` int , `university` varchar(32) NOT NULL, `gpa` float, `active_days_within_30` int , `question_cnt` int , `answer_cnt` int ); CREATE TABLE `question_practice_detail` ( `id` int NOT NULL, `device_id` int NOT NULL, `question_id`int NOT NULL, `result` varchar(32) NOT NULL, `date` date NOT NULL ); CREATE TABLE `question_detail` ( `id` int NOT NULL, `question_id`int NOT NULL, `difficult_level` varchar(32) NOT NULL ); INSERT INTO user_profile VALUES(1,2138,’male’,21,’北京大学’,3.4,7,2,12); INSERT INTO user_profile VALUES(2,3214,’male’,null,’复旦大学’,4.0,15,5,25); INSERT INTO user_profile VALUES(3,6543,’female’,20,’北京大学’,3.2,12,3,30); INSERT INTO user_profile VALUES(4,2315,’female’,23,’浙江大学’,3.6,5,1,2); INSERT INTO user_profile VALUES(5,5432,’male’,25,’山东大学’,3.8,20,15,70); INSERT INTO user_profile VALUES(6,2131,’male’,28,’山东大学’,3.3,15,7,13); INSERT INTO user_profile VALUES(7,4321,’male’,28,’复旦大学’,3.6,9,6,52); INSERT INTO question_practice_detail VALUES(1,2138,111,’wrong’,’2021-05-03′); INSERT INTO question_practice_detail VALUES(2,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(3,3214,113,’wrong’,’2021-06-15′); INSERT INTO question_practice_detail VALUES(4,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(5,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(6,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(7,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(8,3214,112,’wrong’,’2021-05-09′); INSERT INTO question_practice_detail VALUES(9,3214,113,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(10,6543,111,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(11,2315,115,’right’,’2021-08-13′); INSERT INTO question_practice_detail VALUES(12,2315,116,’right’,’2021-08-14′); INSERT INTO question_practice_detail VALUES(13,2315,117,’wrong’,’2021-08-15′); INSERT INTO question_practice_detail VALUES(14,3214,112,’wrong’,’2021-08-16′); INSERT INTO question_practice_detail VALUES(15,3214,113,’wrong’,’2021-08-18′); INSERT INTO question_practice_detail VALUES(16,6543,111,’right’,’2021-08-13′); INSERT INTO question_detail VALUES(1,111,’hard’); INSERT INTO question_detail VALUES(2,112,’medium’); INSERT INTO question_detail VALUES(3,113,’easy’); INSERT INTO question_detail VALUES(4,115,’easy’); INSERT INTO question_detail VALUES(5,116,’medium’); INSERT INTO question_detail VALUES(6,117,’easy’); 输出 easy|0.5000 medium|1.0000 马上挑战 算法知识视频讲解 提交运行 算法知识视频讲解 添加笔记 求解答(0) 邀请回答 收藏(1250) 分享 提交结果有问题? 316个回答 120篇题解 添加回答 17 Mysql Qiuxj SELECT difficult_level, AVG(IF(result=’right’,1,0)) AS correct_rate FROM user_profile u, question_practice_detail qpd, question_detail qd WHERE u.device_id = qpd.device_id AND qpd.question_id = qd.question_id AND university=’浙江大学’ GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-09-29 22:58:04 回复(4) 12 Mysql 老狗丶 select difficult_level, (sum(case when qpd.result = "right" then 1 else 0 end)/count(u.answer_cnt)) as correct_rate from user_profile u inner join question_practice_detail qpd on u.device_id = qpd.device_id inner join question_detail qd on qd.question_id = qpd.question_id where university = "浙江大学" group by difficult_level order by correct_rate; 发表于 2021-08-28 16:11:28 回复(9) 4 智能路障 这题考察的是多表查询，只要把条件梳理到位就可迎刃而解 select difficult_level, count(if(result = ‘right’, 1, null)) / count(result) as correct_rate from user_profile t1, question_practice_detail t2, question_detail t3 where t1.device_id = t2.device_id and t2.question_id = t3.question_id and university = ‘浙江大学’ group by difficult_level order by correct_rate 发表于 2021-10-31 11:17:26 回复(2) 4 fire-light-guo select d.difficult_level, sum(if(result = ‘right’, 1, 0)) / count(*) as correct_rate from user_profile u, question_practice_detail qp, question_detail d where u.university = ‘浙江大学’ and u.device_id = qp.device_id and qp.question_id = d.question_id group by d.difficult_level order by correct_rate 1. 做个三表关联 2. 按问题难度分组 3. sum（）嵌套if（）计算出正确的答题数量，再除以答题总数 发表于 2021-10-13 20:45:23 回复(2) 7 Baily1234 SELECT difficult_level, SUM(IF(result = "wrong", 0, 1)) / count(*) AS correct_rate FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id LEFT JOIN question_detail q2 ON q.question_id = q2.question_id WHERE university = "浙江大学" GROUP BY difficult_level ORDER BY correct_rate; 发表于 2021-10-15 16:22:28 回复(2) 3 茵桃小蚊子 没有上一题难… 现在习惯写代码前开始用加粗关键词和初步处理方法先捋一遍题目，然后就写得很顺手呢。如下： 现在运营想要了解浙江大学(where)的用户在不同难度题目(group by)下答题的正确率(正确数/题数)情况，请取出相应数据，并按照准确率升序(order by asc)输出。 SELECT qd.difficult_level as difficult_level, sum(if(qpd.result="right",1,0))/ COUNT(qpd.result) as correct_rate FROM user_profile u RIGHT JOIN question_practice_detail qpd ON (u.device_id = qpd.device_id) LEFT JOIN question_detail qd ON (qd.question_id = qpd.question_id) WHERE u.university = "浙江大学" GROUP BY qd.difficult_level ORDER BY correct_rate ASC 发表于 2021-11-25 21:05:25 回复(0) 3 Mysql 其实是牛哥 select c.difficult_level, count(if(result=’right’,a.question_id,null)) / count(a.question_id) as correct_rate from (select device_id,question_id,result from question_practice_detail )a join (select device_id from user_profile where university = ‘浙江大学’ )b on a.device_id = b.device_id left join ( select question_id,difficult_level from question_detail )c on a.question_id = c.question_id group by difficult_level order by correct_rate 发表于 2021-08-27 10:57:13 回复(0) 2 Mysql YonChun select difficult_level, count(if(result=’right’,1,null)) / count(result) as correct_rate from user_profile up join question_practice_detail using(device_id) join question_detail using(question_id) where university=’浙江大学’ group by difficult_level order by correct_rate 应该是最精简的了吧 发表于 2022-01-31 11:04:12 回复(0) 2 Mysql 牛客597002486号 select difficult_level, (count(if(result=’right’,1,null)) / count(a.question_id)) as correct_rate from question_practice_detail a join (select device_id from user_profile where university=’浙江大学’) b on a.device_id=b.device_id join question_detail c on a.question_id=c.question_id group by difficult_level order by correct_rate 发表于 2021-10-03 00:47:26 回复(4) 1 Mysql 369186344 select difficult_level, (count(case when t2.result = ‘right’ then 1 else null end)/ count(t2.question_id))correct_rate from (select qp.device_id,question_id,result from question_practice_detail qp where qp.device_id in (select device_id from user_profile up where university = ‘浙江大学’))t2 join question_detail qd on t2.question_id = qd.question_id group by difficult_level order by correct_rate; 发表于 2022-04-22 23:01:00 回复(0) 1 黑白配XY select difficult_level, count(if(result=’right’,result,null))/ count(result) from user_profile u join (select device_id,difficult_level,result from question_practice_detail q1 join question_detail q2 on q1.question_id = q2.question_id) t on u.device_id = t.device_id and university = ‘浙江大学’ group by difficult_level 发表于 2022-04-18 16:25:48 回复(0) 1 Mysql 牛客228167738号 SELECT d.difficult_level, count(if(q.result="right", 1, null)) / count(*) AS correct_rate FROM user_profile AS u INNER JOIN question_practice_detail AS q ON u.device_id = q.device_id INNER JOIN question_detail AS d ON q.question_id = d.question_id WHERE u.university = "浙江大学" GROUP BY d.difficult_level ORDER BY correct_rate ASC; 发表于 2022-04-16 23:45:52 回复(0) 1 Mysql 牛客398096850号 有没有哪位大神帮我看下我的none是哪步错了吗 SELECT Q.difficult_level ,COUNT(CASE WHEN P.result = ‘right’ THEN 1 ELSE NULL END)/COUNT(P.question_id) AS correct_rate FROM ( ( SELECT device_id ,university FROM user_profile WHERE university = ‘浙江大学’ )U LEFT JOIN ( SELECT device_id ,question_id ,result FROM question_practice_detail )P ON P.device_id = U.device_id LEFT JOIN ( SELECT question_id ,difficult_level FROM question_detail )Q ON Q.question_id = P.question_id ) GROUP BY Q.difficult_level ORDER BY correct_rate 发表于 2022-04-05 17:35:03 回复(0) 1 Mysql 柒小漠 select difficult_level, count(case when result =’right’ then 1 else null end)/ count(question_detail.question_id) as rightrate from question_practice_detail join user_profile on question_practice_detail.device_id=user_profile.device_id join question_detail on question_practice_detail.question_id=question_detail.question_id where university = ‘浙江大学’ group by difficult_level order by rightrate 发表于 2022-03-29 14:55:59 回复(0) 1 Mysql Fayuan select t3.difficult_level,sum(case t2.result when ‘right’ then 1 else 0 end) / count(1) as ‘correct_rate’ from user_profile t1 join question_practice_detail t2 on t1.device_id=t2.device_id join question_detail t3 on t2.question_id=t3.question_id where t1.university=’浙江大学’ group by t3.difficult_level order by correct_rate 发表于 2022-03-22 15:07:27 回复(0) 1 XXCC111111 1、学位为浙江大学，为条件；2、不同难度水平的，就表示可能需要根据难度水平来分组，使用group by函数；3、答题正确率用sum(if(qpd.result = ‘right’, 1, 0))/count(*) 来计算 select difficult_level, sum(if(qpd.result = ‘right’, 1, 0))/count(*) as correct_rate from question_practice_detail as qpd left join user_profile as up on qpd.device_id = up.device_id left join question_detail as qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by qd.difficult_level order by correct_rate; 发表于 2022-03-03 13:22:17 回复(0) 1 Mysql 柒鳴飞 select difficult_level, sum(if(result=’right’,1,0))/count(qd.question_id) correct_rate from user_profile up join question_practice_detail qpd on up.device_id = qpd.device_id join question_detail qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by difficult_level order by correct_rate 发表于 2021-12-14 16:48:23 回复(0) 1 Mysql 牛客689071742号 SELECT d.difficult_level, sum( IF ( q.result = ‘right’, 1, 0 ))/ COUNT(*) AS correct_rate FROM question_practice_detail AS q INNER JOIN user_profile AS u ON u.device_id = q.device_id AND u.university = ‘浙江大学’ LEFT JOIN question_detail AS d ON q.question_id = d.question_id GROUP BY d.difficult_level ORDER BY correct_rate ASC 发表于 2021-11-20 11:22:56 回复(0) 1 Mysql 键盘盘盘盘 select difficult_level, sum(case when result = ‘right’ then 1 else 0 end)/count(*) as correct_rate from question_practice_detail as qpd left join user_profile as u on u.device_id = qpd.device_id left join question_detail as qd on qpd.question_id = qd.question_id where university = ‘浙江大学’ group by qd.difficult_level order by correct_rate 1注意写的时候如果都和中间的表有关,就得让中间的表left join其他的两个表 2注意count是统计行数,并不是计算,如果要计算的话,需要使用sum函数计算列里面的值!! 发表于 2021-11-08 14:50:46 回复(0) 1 Mysql 码农矿工小付 SELECT tmp.difficult_level, AVG(IF(tmp.result = ‘right’, 1, 0)) AS correct_rate FROM (SELECT qpd.question_id, qpd.result, qd.difficult_level, qpd.date FROM user_profile AS up LEFT JOIN question_practice_detail AS qpd ON up.device_id = qpd.device_id LEFT JOIN question_detail AS qd ON qpd.question_id = qd.question_id WHERE up.university = ‘浙江大学’) AS tmp GROUP BY tmp.difficult_level; 这个怎么改啊？用例就是一个过不去，而且是顺序问题，这种该怎么处理啊？ 发表于 2021-08-25 15:14:02 回复(7) 这道题你会答吗？花几分钟告诉大家答案吧！ 提交观点
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