这适用于SQL 2000:

DECLARE @testTable TABLE 
( 
    VALUE   INT
)
--INSERT INTO @testTable -- Even Test
--SELECT 3 UNION ALL
--SELECT 5 UNION ALL
--SELECT 7 UNION ALL
--SELECT 12 UNION ALL
--SELECT 13 UNION ALL
--SELECT 14 UNION ALL
--SELECT 21 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 29 UNION ALL
--SELECT 40 UNION ALL
--SELECT 56

--
--INSERT INTO @testTable -- Odd Test
--SELECT 3 UNION ALL
--SELECT 5 UNION ALL
--SELECT 7 UNION ALL
--SELECT 12 UNION ALL
--SELECT 13 UNION ALL
--SELECT 14 UNION ALL
--SELECT 21 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 23 UNION ALL
--SELECT 29 UNION ALL
--SELECT 39 UNION ALL
--SELECT 40 UNION ALL
--SELECT 56


DECLARE @RowAsc TABLE
(
    ID      INT IDENTITY,
    Amount  INT
)

INSERT INTO @RowAsc
SELECT  VALUE 
FROM    @testTable 
ORDER BY VALUE ASC

SELECT  AVG(amount)
FROM @RowAsc ra
WHERE ra.id IN
(
    SELECT  ID 
    FROM    @RowAsc
    WHERE   ra.id -
    (
        SELECT  MAX(id) / 2.0 
        FROM    @RowAsc
    ) BETWEEN 0 AND 1

)

对于像我这样正在学习基础知识的新手来说，我个人觉得这个例子更容易理解，因为它更容易理解到底发生了什么以及中值来自哪里……

select
 ( max(a.[Value1]) + min(a.[Value1]) ) / 2 as [Median Value1]
,( max(a.[Value2]) + min(a.[Value2]) ) / 2 as [Median Value2]

from (select
    datediff(dd,startdate,enddate) as [Value1]
    ,xxxxxxxxxxxxxx as [Value2]
     from dbo.table1
     )a

不过，对上面的一些代码绝对敬畏!!

我只是在寻找一个基于集的中位数的解决方案时偶然发现了这一页。在研究了一些解决方案之后，我想到了以下几点。希望是有用的。

DECLARE @test TABLE(
    i int identity(1,1),
    id int,
    score float
)

INSERT INTO @test (id,score) VALUES (1,10)
INSERT INTO @test (id,score) VALUES (1,11)
INSERT INTO @test (id,score) VALUES (1,15)
INSERT INTO @test (id,score) VALUES (1,19)
INSERT INTO @test (id,score) VALUES (1,20)

INSERT INTO @test (id,score) VALUES (2,20)
INSERT INTO @test (id,score) VALUES (2,21)
INSERT INTO @test (id,score) VALUES (2,25)
INSERT INTO @test (id,score) VALUES (2,29)
INSERT INTO @test (id,score) VALUES (2,30)

INSERT INTO @test (id,score) VALUES (3,20)
INSERT INTO @test (id,score) VALUES (3,21)
INSERT INTO @test (id,score) VALUES (3,25)
INSERT INTO @test (id,score) VALUES (3,29)

DECLARE @counts TABLE(
    id int,
    cnt int
)

INSERT INTO @counts (
    id,
    cnt
)
SELECT
    id,
    COUNT(*)
FROM
    @test
GROUP BY
    id

SELECT
    drv.id,
    drv.start,
    AVG(t.score)
FROM
    (
        SELECT
            MIN(t.i)-1 AS start,
            t.id
        FROM
            @test t
        GROUP BY
            t.id
    ) drv
    INNER JOIN @test t ON drv.id = t.id
    INNER JOIN @counts c ON t.id = c.id
WHERE
    t.i = ((c.cnt+1)/2)+drv.start
    OR (
        t.i = (((c.cnt+1)%2) * ((c.cnt+2)/2))+drv.start
        AND ((c.cnt+1)%2) * ((c.cnt+2)/2) <> 0
    )
GROUP BY
    drv.id,
    drv.start

虽然Justin grant的解决方案看起来很可靠，但我发现当您在给定的分区键中有许多重复值时，ASC重复值的行号最终会不按顺序排列，因此它们不能正确对齐。

以下是我的研究结果的一个片段:

KEY VALUE ROWA ROWD  

13  2     22   182
13  1     6    183
13  1     7    184
13  1     8    185
13  1     9    186
13  1     10   187
13  1     11   188
13  1     12   189
13  0     1    190
13  0     2    191
13  0     3    192
13  0     4    193
13  0     5    194

我使用Justin的代码作为这个解决方案的基础。尽管考虑到使用多个派生表效率不高，但它确实解决了我遇到的行排序问题。任何改进都会受到欢迎，因为我在T-SQL方面不是那么有经验。

SELECT PKEY, cast(AVG(VALUE)as decimal(5,2)) as MEDIANVALUE
FROM
(
  SELECT PKEY,VALUE,ROWA,ROWD,
  'FLAG' = (CASE WHEN ROWA IN (ROWD,ROWD-1,ROWD+1) THEN 1 ELSE 0 END)
  FROM
  (
    SELECT
    PKEY,
    cast(VALUE as decimal(5,2)) as VALUE,
    ROWA,
    ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY ROWA DESC) as ROWD 

    FROM
    (
      SELECT
      PKEY, 
      VALUE,
      ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY VALUE ASC,PKEY ASC ) as ROWA 
      FROM [MTEST]
    )T1
  )T2
)T3
WHERE FLAG = '1'
GROUP BY PKEY
ORDER BY PKEY

以下解决方案在这些假设下有效:

无重复值 没有取消

代码:

IF OBJECT_ID('dbo.R', 'U') IS NOT NULL
  DROP TABLE dbo.R

CREATE TABLE R (
    A FLOAT NOT NULL);

INSERT INTO R VALUES (1);
INSERT INTO R VALUES (2);
INSERT INTO R VALUES (3);
INSERT INTO R VALUES (4);
INSERT INTO R VALUES (5);
INSERT INTO R VALUES (6);

-- Returns Median(R)
select SUM(A) / CAST(COUNT(A) AS FLOAT)
from R R1 
where ((select count(A) from R R2 where R1.A > R2.A) = 
      (select count(A) from R R2 where R1.A < R2.A)) OR
      ((select count(A) from R R2 where R1.A > R2.A) + 1 = 
      (select count(A) from R R2 where R1.A < R2.A)) OR
      ((select count(A) from R R2 where R1.A > R2.A) = 
      (select count(A) from R R2 where R1.A < R2.A) + 1) ; 

在Jeff Atwood的答案的基础上，它是用GROUP BY和一个相关的子查询来获得每个组的中位数。

SELECT TestID, 
(
 (SELECT MAX(Score) FROM
   (SELECT TOP 50 PERCENT Score FROM Posts WHERE TestID = Posts_parent.TestID ORDER BY Score) AS BottomHalf)
 +
 (SELECT MIN(Score) FROM
   (SELECT TOP 50 PERCENT Score FROM Posts WHERE TestID = Posts_parent.TestID ORDER BY Score DESC) AS TopHalf)
) / 2 AS MedianScore,
AVG(Score) AS AvgScore, MIN(Score) AS MinScore, MAX(Score) AS MaxScore
FROM Posts_parent
GROUP BY Posts_parent.TestID