• sql之表的表达式


    1、派生表

    实质:就是特殊的子查询(将查询结果放在from后面)

    含有一张Order表:

    看下面的sql语句:

    1 select
    2 orderid,orderdate,custid
    3 from
    4 (
    5    select
    6    orderid,orderdate,custid,ROW_NUMBER() over(order by orderid) as rownum--列明必须起别名
    7    from  [Sales.Orders]
    8 ) as t--表名必须起别名

    需要注意的特殊之处:查询出来的表要起别名,子查询里面的字段名也要起别名

    2、CTE(公共表的表达式)

     1)语法规范:

    1 with use_country--定义的表名
    2 as--上面的是 语法:
    3 (
    4     select country,companyname,custid
    5     from [Sales.Customers]
    6     where country='按时打算'
    7 )
    8 select * from use_country;--将查询语句 写在 子查询的后面

    将查询语句放在最后。

    下面看几个例子,具体解析一下:

    业务逻辑:还是用上面的那张Order表,要求:查询出 每年的订单数量 大于10的 用户

    1》普通的写法:

    1 select
    2 custid,year(orderdate),COUNT(1) as '订单数量'
    3 from [Sales.Orders]
    4 group by YEAR(orderdate), custid
    5 having COUNT(orderid)>10--having:对 group by  之后的 组函数 进行 筛选使用

     2》使用派生表:

     1 select
     2 custid,orderyear,orderNum
     3 from
     4 (
     5     select
     6     custid,orderyear,COUNT(*) as orderNum
     7     from
     8     (
     9         select
    10         custid, year(orderdate) as orderyear
    11         from [Sales.Orders]
    12     ) as t1
    13     group by orderyear,custid
    14 ) as t2
    15 where orderNum>10

    上面的 查询的层次只有两层,如果层次非常多的话,就会出现,看起来非常费劲了,所以使用CTE的方式更简洁

     1 with OrderYear
     2 as
     3 (
     4     select
     5     custid,YEAR(orderdate) as orderyear
     6     from [Sales.Orders]
     7 ),
     8 OrderGroupYear
     9 as
    10 (
    11   select
    12   custid,orderyear,COUNT(orderyear) as ordernum
    13   from OrderYear
    14   group by orderyear,custid
    15 ),
    16 OrderNumThanTen
    17 as
    18 (
    19   select
    20   custid,orderyear,ordernum
    21   from OrderGroupYear
    22   where ordernum>10
    23 )
    24 select * from OrderNumThanTen;

    2)CTE还可以多张表引用:(相当于c#里面的将重复的代码封装成一个方法)下面举例:

    业务逻辑:查询出 每年 客户的数量,以及 前后两年之间客户数量的差量

    --首先查询出每年叫客户的数量:

    1 select
    2 year(orderdate),COUNT(distinct custid)--将 重复的 客户 要去掉
    3 from [Sales.Orders]
    4 group by YEAR(orderdate)

    --然后使用派生表的方式实现:

     1 select
     2 currtYearOrder.orderYear,currtYearOrder.custCount,prevYearOrder.orderYear,prevYearOrder.custCount,currtYearOrder.custCount-prevYearOrder.custCount
     3 from
     4 (
     5     select
     6     year(orderdate) as orderYear,COUNT(distinct custid) as custCount--将 重复的 客户 要去掉
     7     from [Sales.Orders]
     8     group by YEAR(orderdate)
     9 ) as currtYearOrder
    10 left outer join 
    11 (
    12     select
    13     year(orderdate) as orderYear ,COUNT(distinct custid) as custCount--将 重复的 客户 要去掉
    14     from [Sales.Orders]
    15     group by YEAR(orderdate)
    16 ) as prevYearOrder
    17 on currtYearOrder.orderYear=prevYearOrder.orderYear+1

    --下面使用CTE的方式实现,就不用重复:

     1 with OrderYearCust
     2 as
     3 (
     4     select
     5     year(orderdate) as orderYear,COUNT(distinct custid) as custCount--将 重复的 客户 要去掉
     6     from [Sales.Orders]
     7     group by YEAR(orderdate)
     8 )
     9 select 
    10 curtYearOrder.orderYear,curtYearOrder.custCount,prevYearOrder.orderYear,prevYearOrder.custCount,curtYearOrder.custCount-prevYearOrder.custCount
    11 from OrderYearCust as curtYearOrder left outer join OrderYearCust as prevYearOrder on curtYearOrder.orderYear=prevYearOrder.orderYear+1

    3)CTE递归查询:(针对 树形节点 进行查询)

    表的结果如下:

    业务逻辑:需要查询出 mgrid 为 2 下面的 所有的子节点

     1 with diguiEmployee
     2 as
     3 (
     4    --起点:最上层的 查询(只执行一次)
     5     select 
     6     enpid,lastname,firstname,mgrid
     7     from [HR.Employees]
     8     where mgrid=6
     9    
    10     union all --连接 起点:上层查询 和递归查询
    11     
    12     --递归查询
    13     select 
    14     e.enpid,e.lastname,e.firstname,e.mgrid
    15     from [HR.Employees] as e inner  join diguiEmployee as d
    16     on e.mgrid=d.enpid
    17 )
    18 select     enpid,lastname,firstname,mgrid from [HR.Employees] where enpid=6
    19 union all --将查询的 结果 连接 起来
    20 select * from diguiEmployee;
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  • 原文地址:https://www.cnblogs.com/xiaoxiaogogo/p/3440534.html
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