Mysql中表student_table(id,name,birth,sex),插入如下记录: (‘1004’ , ‘张三’ , ‘2000-08-06’ , ‘男’); (‘1005’ , NULL , ‘2001-12-01’ , ‘女’); (‘1006’ , ‘张三’ , ‘2000-08-06’ , ‘女’); (‘1007’ , ‘王五’ , ‘2001-12-01’ , ‘男’); (‘1008’ , ‘李四’ , NULL, ‘女’); (‘1009’ , ‘李四’ , NULL, ‘男’); (‘1010’ , ‘李四’ , ‘2001-12-01’, ‘女’); 执行 select count(t1.birth) as c1 from ( select * from student_table where sex = ‘男’ ) t1  right  join  (select * from student_table where sex = ‘女’) t2  on t1.birth = t2.birth and t1.name = t2.name ;  的结果行数是()?

区块链毕设网qklbishe.com为您提供问题的解答

Mysql中表student_table(id,name,birth,sex),插入如下记录:

(‘1004’ , ‘张三’ , ‘2000-08-06’ , ‘男’);

(‘1005’ , NULL , ‘2001-12-01’ , ‘女’);

(‘1006’ , ‘张三’ , ‘2000-08-06’ , ‘女’);

(‘1007’ , ‘王五’ , ‘2001-12-01’ , ‘男’);

(‘1008’ , ‘李四’ , NULL, ‘女’);

(‘1009’ , ‘李四’ , NULL, ‘男’);

(‘1010’ , ‘李四’ , ‘2001-12-01’, ‘女’);

执行

select count(t1.birth) as c1

from (

select * from student_table where sex = ‘男’ ) t1 

right  join 

(select * from student_table where sex = ‘女’) t2 

on t1.birth = t2.birth and t1.name = t2.name ; 

的结果行数是()?

NULL 和 Null不匹配吗
27:24

以上就是关于问题Mysql中表student_table(id,name,birth,sex),插入如下记录: (‘1004’ , ‘张三’ , ‘2000-08-06’ , ‘男’); (‘1005’ , NULL , ‘2001-12-01’ , ‘女’); (‘1006’ , ‘张三’ , ‘2000-08-06’ , ‘女’); (‘1007’ , ‘王五’ , ‘2001-12-01’ , ‘男’); (‘1008’ , ‘李四’ , NULL, ‘女’); (‘1009’ , ‘李四’ , NULL, ‘男’); (‘1010’ , ‘李四’ , ‘2001-12-01’, ‘女’); 执行 select count(t1.birth) as c1 from ( select * from student_table where sex = ‘男’ ) t1  right  join  (select * from student_table where sex = ‘女’) t2  on t1.birth = t2.birth and t1.name = t2.name ;  的结果行数是()?的答案

欢迎关注区块链毕设网-
专业区块链毕业设计成品源码,定制。

区块链NFT链游项目方科学家脚本开发培训

从业7年-专注一级市场


微信:btc9767
TELEGRAM :https://t.me/btcok9

具体资料介绍

web3的一级市场千万收益的逻辑


进群点我



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 » Mysql中表student_table(id,name,birth,sex),插入如下记录: (‘1004’ , ‘张三’ , ‘2000-08-06’ , ‘男’); (‘1005’ , NULL , ‘2001-12-01’ , ‘女’); (‘1006’ , ‘张三’ , ‘2000-08-06’ , ‘女’); (‘1007’ , ‘王五’ , ‘2001-12-01’ , ‘男’); (‘1008’ , ‘李四’ , NULL, ‘女’); (‘1009’ , ‘李四’ , NULL, ‘男’); (‘1010’ , ‘李四’ , ‘2001-12-01’, ‘女’); 执行 select count(t1.birth) as c1 from ( select * from student_table where sex = ‘男’ ) t1  right  join  (select * from student_table where sex = ‘女’) t2  on t1.birth = t2.birth and t1.name = t2.name ;  的结果行数是()?