SQL查询，按外部键排序按日期排序前2 2

Question

我需要一个SQL查询，它返回每个ClientID的PlanDate的前2个计划。这是所有在PlanID是PrimaryID，ClientID是一个foreignID。

这是我迄今为止 - >

SELECT * 
FROM [dbo].[tblPlan] 
WHERE [PlanID] IN (SELECT TOP (2) PlanID FROM [dbo].[tblPlan] ORDER BY [PlanDate] DESC) 

显然，这只能返回2条记录，我确实需要最多2个记录每ClientID。

Answer 1

这可以通过使用ROW_NUMBER来完成：

SELECT PlanId, ClientId, PlanDate FROM (
    SELECT ROW_NUMBER() OVER (PARTITION BY ClientId ORDER BY PlanDate DESC) rn, * 
    FROM [dbo].[tblPlan] 
) AS T1 
WHERE rn <=2 

添加任何其他列你需要选择以获得这些。

Answer 2

编辑，12月2011.更正CROSS应用解决方案

尝试既看到什么是最好的

SELECT * 
FROM 
    (-- distinct ClientID values 
    SELECT DISTINCT ClientID 
    FROM [dbo].[tblPlan] 
) P1 
    CROSS APPLY 
    (-- top 2 per ClientID 
    SELECT TOP (2) P2.PlanID 
    FROM [dbo].[tblPlan] P2 
    WHERE P1.ClientID = P2.ClientID 
    ORDER BY P2.[PlanDate] DESC 
) foo 

或者

;WITH cTE AS (
    SELECT 
    *, 
    ROW_NUMBER() OVER (PARTITION BY clientid ORDER BY [PlanDate] DESC) AS Ranking 
    FROM 
    [dbo].[tblPlan] 
) 
SELECT * FROM cTE WHERE Ranking <= 2