SQL查詢，按外部鍵排序按日期排序前2 2

Question

我需要一個SQL查詢，它返回每個ClientID的PlanDate的前2個計劃。這是所有在PlanID是PrimaryID，ClientID是一個foreignID。

這是我迄今爲止 - >

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

顯然，這隻能返回2條記錄，我確實需要最多2個記錄每ClientID。

Answer 1

這可以通過使用ROW_NUMBER來完成：

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選擇以獲得這些。

Answer 2

編輯，12月2011.更正CROSS應用解決方案

嘗試既看到什麼是最好的

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