如何在使用WITH CUBE時獲得中位數值？

Question

簡而言之：在MSSQL 2005中使用WITH CUBE時，是否有任何方法直接計算中間值，最好不使用循環？

一些代碼：

CREATE TABLE #Example (
    ID1 INT, 
    ID2 INT, 
    [Value] INT 
) 

INSERT #Example(ID1,ID2,[Value]) 
    SELECT 1, 1, 45 UNION 
    SELECT 1, 2, 78 UNION 
    SELECT 1, 3, 23 UNION 
    SELECT 1, 4, 119 UNION 
    SELECT 1, 5, 85 UNION 
    SELECT 2, 1, 143 UNION 
    SELECT 2, 2, 124 UNION 
    SELECT 2, 3, 161 UNION 
    SELECT 2, 4, 171 UNION 
    SELECT 2, 5, 142 

SELECT ID1, ID2, COUNT(1) AS 'NumValues', MIN([Value]) AS 'Minimum', AVG([Value]) AS 'Average', MAX([Value]) AS 'Maximum' 
    FROM #Example 
    GROUP BY ID1, ID2 
    WITH CUBE 
    HAVING ID1 IS NULL OR ID2 IS NULL 
    ORDER BY ISNULL(ID1, 0), ISNULL(ID2, 0) 

DROP TABLE #Example 

我如何獲得平均值的列？

迄今爲止管理得最好的是循環ID1和ID2值，並計算每組ID的中位數，但這似乎是低效和醜陋的。

任何想法感激地收到！

Answer 1

有點爲時已晚，我認爲這會工作：

select * from 
(
    SELECT ID1, ID2, COUNT(1) AS 'NumValues', MIN([Value]) AS 'Minimum', AVG([Value]) AS 'Average', MAX([Value]) AS 'Maximum' 
    FROM #Example 
    GROUP BY ID1, ID2 
    WITH CUBE 
    HAVING ID1 IS NULL OR ID2 IS NULL 
) T 
OUTER APPLY 
(
    select avg(val) as median from 
    (
     select [value] as val, row_number() over (order by [value]) as rn 
     from #Example E 
     where (T.ID1=E.ID1 or T.ID1 is null) and (T.ID2=E.ID2 or T.ID2 is null) 
    ) I 
    where abs(rn - (T.NumValues+1)/2.0) <1 
) T2