有两张数据表存在于SQLServer中定义如下: CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), ) 以下的SQL-查询来自上海且订单少于3个的客户以及他们的订单数,查询结果按订单数从小到大的顺序排列 ① select c.customerid,count(o.orderid)as numorders ② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid ③ where C.city=’Shanghai’ ④ group by c.customerid ⑤ having count(O.orderid)<3 ⑥ order by numorders 该SQL逻辑查询处理顺序应该是
区块链毕设网qklbishe.com为您提供问题的解答
CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), )
① select c.customerid,count(o.orderid)as numorders
② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid
③ where C.city=’Shanghai’
④ group by c.customerid
⑤ having count(O.orderid)<3
⑥ order by numorders
该SQL逻辑查询处理顺序应该是
结构:查询语句+group by+ having +聚合函数统计
直接上栗子:
SELECT emp_no, count(salary) t # 在工资表里查找员工编号、统计发工资的次数(取别名t) from salaries # 工资表 where creat_date between '2020-01-01' and '2020-12-31' group by emp_no # 通过员工编号编组 HAVING t>15 # 查询发工资次数大于15次的
常用聚合函数是sum()、avg()、count();
(聚合函数是用来统计每个分组的统计信息,它们要跟 group by 一起使用,用来将每个分组所有数据聚合成一条统计数据。聚合函数: max() min() count() avg() sum() 等。
语句执行的顺序
select 语句执行顺序大致是:
where(数据查询) -> group by(数据编组) -> having(结果过滤) -> order by(排序)
以上就是关于问题有两张数据表存在于SQLServer中定义如下: CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), ) 以下的SQL-查询来自上海且订单少于3个的客户以及他们的订单数,查询结果按订单数从小到大的顺序排列 ① select c.customerid,count(o.orderid)as numorders ② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid ③ where C.city=’Shanghai’ ④ group by c.customerid ⑤ having count(O.orderid)<3 ⑥ order by numorders 该SQL逻辑查询处理顺序应该是的答案
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 » 有两张数据表存在于SQLServer中定义如下: CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), ) 以下的SQL-查询来自上海且订单少于3个的客户以及他们的订单数,查询结果按订单数从小到大的顺序排列 ① select c.customerid,count(o.orderid)as numorders ② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid ③ where C.city=’Shanghai’ ④ group by c.customerid ⑤ having count(O.orderid)<3 ⑥ order by numorders 该SQL逻辑查询处理顺序应该是
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 » 有两张数据表存在于SQLServer中定义如下: CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), ) 以下的SQL-查询来自上海且订单少于3个的客户以及他们的订单数,查询结果按订单数从小到大的顺序排列 ① select c.customerid,count(o.orderid)as numorders ② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid ③ where C.city=’Shanghai’ ④ group by c.customerid ⑤ having count(O.orderid)<3 ⑥ order by numorders 该SQL逻辑查询处理顺序应该是
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 » 有两张数据表存在于SQLServer中定义如下: CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), ) 以下的SQL-查询来自上海且订单少于3个的客户以及他们的订单数,查询结果按订单数从小到大的顺序排列 ① select c.customerid,count(o.orderid)as numorders ② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid ③ where C.city=’Shanghai’ ④ group by c.customerid ⑤ having count(O.orderid)<3 ⑥ order by numorders 该SQL逻辑查询处理顺序应该是
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 » 有两张数据表存在于SQLServer中定义如下: CREATE TABLE [dbo].[Customers]( [customerid] [char](5) NOT NULL PRIMARY KEY , [city] [varchar](10) NOT NULL, ) CREATE TABLE [dbo].[Orders]( [orderid] [int] NOT NULL PRIMARY KEY, [customerid] [char](5) NULL references Customers(customerid), ) 以下的SQL-查询来自上海且订单少于3个的客户以及他们的订单数,查询结果按订单数从小到大的顺序排列 ① select c.customerid,count(o.orderid)as numorders ② from [Customers] as C left outer join [Orders] as O on C.customerid=O.customerid ③ where C.city=’Shanghai’ ④ group by c.customerid ⑤ having count(O.orderid)<3 ⑥ order by numorders 该SQL逻辑查询处理顺序应该是