如何组合 GROUP BY 和 ROW_NUMBER？答案

【问题标题】：How to combine GROUP BY and ROW_NUMBER?如何组合 GROUP BY 和 ROW_NUMBER？
【发布时间】：2012-03-25 22:24:09
【问题描述】：

我希望下面的示例代码是不言自明的：

declare @t1 table (ID int,Price money, Name varchar(10))
declare @t2 table (ID int,Orders int,  Name varchar(10))
declare @relation  table (t1ID int,t2ID int)
insert into @t1 values(1, 200, 'AAA');
insert into @t1 values(2, 150, 'BBB');
insert into @t1 values(3, 100, 'CCC');
insert into @t2 values(1,25,'aaa');
insert into @t2 values(2,35,'bbb');
insert into @relation values(1,1);
insert into @relation values(2,1);
insert into @relation values(3,2);

select T2.ID AS T2ID
,T2.Name as T2Name
,T2.Orders
,T1.ID AS T1ID
,T1.Name As T1Name
,T1Sum.Price
FROM @t2 T2
INNER JOIN (
    SELECT Rel.t2ID
        ,MAX(Rel.t1ID)AS t1ID 
-- the MAX returns an arbitrary ID, what i need is: 
--      ,ROW_NUMBER()OVER(Partition By Rel.t2ID Order By Price DESC)As PriceList
        ,SUM(Price)AS Price
        FROM @t1 T1 
        INNER JOIN @relation Rel ON Rel.t1ID=T1.ID
        GROUP BY Rel.t2ID
)AS T1Sum ON  T1Sum.t2ID = T2.ID
INNER JOIN @t1 T1 ON T1Sum.t1ID=T1.ID

结果：

T2ID   T2Name   Orders  T1ID    T1Name  Price     
 1      aaa       25     2       BBB    350,00     
 2      bbb       35     3       CCC    100,00

上面评论了我需要的东西，这是一种获取ROW_NUMBER 的方法，但也是首先获取Group By 的方法。所以我需要在关系表中按T2.ID 分组的所有T1 价格中的sum，在外部查询中需要最高价格的t1ID。

换句话说：如何将MAX(Rel.t1ID)AS t1ID 更改为返回价格最高的ID？

所以想要的结果是（注意第一个 T1ID 从 2 变为 1，因为它的价格更高）：

T2ID   T2Name   Orders  T1ID    T1Name  Price     
 1      aaa       25     1       AAA    350,00     
 2      bbb       35     3       CCC    100,00

注意：如果您想知道为什么我不将 Orders 与价格相乘：它们不是真实的（所以我应该离开这个专栏，因为它有点模棱两可，请忽略它，我刚刚添加它以减少抽象）。实际上Orders 必须保持不变，这就是子查询方法加入两者的原因，也是我首先需要分组的原因。

结论：显然我的问题的核心可以通过OVER clause 来回答，该OVER clause 可以应用于任何聚合函数，如SUM（见Damien's answer ) 什么对我来说是新的。谢谢大家的工作方法。

【问题讨论】：

最终结果中不应该是AAA而不是BBB吗？

标签： sql sql-server-2005 tsql group-by

【解决方案1】：

哇，其他答案看起来很复杂 - 所以我希望我没有遗漏一些明显的东西。

您可以对聚合使用OVER/PARTITION BY，然后它们将在没有GROUP BY 子句的情况下进行分组/聚合。所以我只是将您的查询修改为：

select T2.ID AS T2ID
    ,T2.Name as T2Name
    ,T2.Orders
    ,T1.ID AS T1ID
    ,T1.Name As T1Name
    ,T1Sum.Price
FROM @t2 T2
INNER JOIN (
    SELECT Rel.t2ID
        ,Rel.t1ID
 --       ,MAX(Rel.t1ID)AS t1ID 
-- the MAX returns an arbitrary ID, what i need is: 
      ,ROW_NUMBER()OVER(Partition By Rel.t2ID Order By Price DESC)As PriceList
        ,SUM(Price)OVER(PARTITION BY Rel.t2ID) AS Price
        FROM @t1 T1 
        INNER JOIN @relation Rel ON Rel.t1ID=T1.ID
--        GROUP BY Rel.t2ID
)AS T1Sum ON  T1Sum.t2ID = T2.ID
INNER JOIN @t1 T1 ON T1Sum.t1ID=T1.ID
where t1Sum.PriceList = 1

它给出了请求的结果集。

【讨论】：

哇，这就是我所希望的，而其他答案正是我所担心的。谢谢（全部），我必须仔细看看如何将它安装到我的实际查询中（我的样本非常减少了）。
谢谢！我一直在试图弄清楚如何使它适用于几个不同的窗口函数。我一直把总和放在 partition by 子句中，而不是放在开头！

【解决方案2】：

毫无疑问，这可以简化，但结果符合您的期望。

主要是为了

在单独的CTE 中计算每个t2ID 的最高价格
在单独的CTE 中计算每个t2ID 的总价
结合CTE的两个结果

SQL 语句

;WITH MaxPrice AS ( 
    SELECT  t2ID
            , t1ID
    FROM    (       
                SELECT  t2.ID AS t2ID
                        , t1.ID AS t1ID
                        , rn = ROW_NUMBER() OVER (PARTITION BY t2.ID ORDER BY t1.Price DESC)
                FROM    @t1 t1
                        INNER JOIN @relation r ON r.t1ID = t1.ID        
                        INNER JOIN @t2 t2 ON t2.ID = r.t2ID
            ) maxt1
    WHERE   maxt1.rn = 1                            
)
, SumPrice AS (
    SELECT  t2ID = t2.ID
            , Price = SUM(Price)
    FROM    @t1 t1
            INNER JOIN @relation r ON r.t1ID = t1.ID
            INNER JOIN @t2 t2 ON t2.ID = r.t2ID
    GROUP BY
            t2.ID           
)           
SELECT  t2.ID
        , t2.Name
        , t2.Orders
        , mp.t1ID
        , t1.ID
        , t1.Name
        , sp.Price
FROM    @t2 t2
        INNER JOIN MaxPrice mp ON mp.t2ID = t2.ID
        INNER JOIN SumPrice sp ON sp.t2ID = t2.ID
        INNER JOIN @t1 t1 ON t1.ID = mp.t1ID

【讨论】：

【解决方案3】：

;with C as
(
  select Rel.t2ID,
         Rel.t1ID,
         t1.Price,
         row_number() over(partition by Rel.t2ID order by t1.Price desc) as rn
  from @t1 as T1
    inner join @relation as Rel
      on T1.ID = Rel.t1ID
)
select T2.ID as T2ID,
       T2.Name as T2Name,
       T2.Orders,
       T1.ID as T1ID,
       T1.Name as T1Name,
       T1Sum.Price
from @t2 as T2
  inner join (
              select C1.t2ID,
                     sum(C1.Price) as Price,
                     C2.t1ID
              from C as C1
                inner join C as C2 
                  on C1.t2ID = C2.t2ID and
                     C2.rn = 1
              group by C1.t2ID, C2.t1ID
             ) as T1Sum
    on T2.ID = T1Sum.t2ID
  inner join @t1 as T1
    on T1.ID = T1Sum.t1ID

【讨论】：

【解决方案4】：

重复数据删除（选择最大 T1）和聚合需要作为不同的步骤进行。我使用了 CTE，因为我认为这更清楚：

;WITH sumCTE
AS
(
    SELECT  Rel.t2ID, SUM(Price) price
    FROM    @t1         AS T1
    JOIN    @relation   AS Rel 
    ON      Rel.t1ID=T1.ID
    GROUP 
    BY      Rel.t2ID
)
,maxCTE
AS
(
    SELECT  Rel.t2ID, Rel.t1ID, 
            ROW_NUMBER()OVER(Partition By Rel.t2ID Order By Price DESC)As PriceList
    FROM    @t1         AS T1
    JOIN    @relation   AS Rel 
    ON      Rel.t1ID=T1.ID
)
SELECT T2.ID AS T2ID
,T2.Name as T2Name
,T2.Orders
,T1.ID AS T1ID
,T1.Name As T1Name
,sumT1.Price
FROM    @t2 AS T2
JOIN    sumCTE AS sumT1
ON      sumT1.t2ID = t2.ID
JOIN    maxCTE AS maxT1
ON      maxT1.t2ID = t2.ID
JOIN    @t1 AS T1
ON      T1.ID = maxT1.t1ID
WHERE   maxT1.PriceList = 1

【讨论】：