使用一条语句——一种方法是使用ROW_NUMBER()， COUNT()窗口函数并过滤子查询。下面是薪资中位数:

 SELECT AVG(e_salary) 
 FROM                                                             
    (SELECT 
      ROW_NUMBER() OVER(ORDER BY e_salary) as row_no, 
      e_salary,
      (COUNT(*) OVER()+1)*0.5 AS row_half
     FROM Employee) t
 WHERE row_no IN (FLOOR(row_half),CEILING(row_half))

我在网上看到过类似的解决方案，使用地板和天花板，但尝试使用单一的语句。(编辑)

中找到

这是查找属性中值的最简单方法。

Select round(S.salary,4) median from employee S 
where (select count(salary) from station 
where salary < S.salary ) = (select count(salary) from station
where salary > S.salary)

我想自己想出一个解决办法，但我的大脑绊倒了。我觉得很管用，但别让我早上解释。: P

DECLARE @table AS TABLE
(
    Number int not null
);

insert into @table select 2;
insert into @table select 4;
insert into @table select 9;
insert into @table select 15;
insert into @table select 22;
insert into @table select 26;
insert into @table select 37;
insert into @table select 49;

DECLARE @Count AS INT
SELECT @Count = COUNT(*) FROM @table;

WITH MyResults(RowNo, Number) AS
(
    SELECT RowNo, Number FROM
        (SELECT ROW_NUMBER() OVER (ORDER BY Number) AS RowNo, Number FROM @table) AS Foo
)
SELECT AVG(Number) FROM MyResults WHERE RowNo = (@Count+1)/2 OR RowNo = ((@Count+1)%2) * ((@Count+2)/2)

2019 UPDATE: In the 10 years since I wrote this answer, more solutions have been uncovered that may yield better results. Also, SQL Server releases since then (especially SQL 2012) have introduced new T-SQL features that can be used to calculate medians. SQL Server releases have also improved its query optimizer which may affect perf of various median solutions. Net-net, my original 2009 post is still OK but there may be better solutions on for modern SQL Server apps. Take a look at this article from 2012 which is a great resource: https://sqlperformance.com/2012/08/t-sql-queries/median

本文发现，以下模式比所有其他选择都要快得多，至少在他们测试的简单模式上是这样。该解决方案比测试的最慢解决方案(PERCENTILE_CONT)快373x (!!)注意，这个技巧需要两个独立的查询，这可能不是在所有情况下都可行。它还需要SQL 2012或更高版本。

DECLARE @c BIGINT = (SELECT COUNT(*) FROM dbo.EvenRows);

SELECT AVG(1.0 * val)
FROM (
    SELECT val FROM dbo.EvenRows
     ORDER BY val
     OFFSET (@c - 1) / 2 ROWS
     FETCH NEXT 1 + (1 - @c % 2) ROWS ONLY
) AS x;

当然，仅仅因为2012年对一个模式的一次测试产生了很好的结果，您的实际情况可能会有所不同，特别是如果您使用的是SQL Server 2014或更高版本。如果性能对中值计算很重要，我强烈建议尝试并测试那篇文章中推荐的几个选项，以确保您找到了最适合您的模式的选项。

我还会特别小心地使用(SQL Server 2012新增的)函数PERCENTILE_CONT，这是这个问题的其他答案之一中推荐的，因为上面链接的文章发现这个内置函数比最快的解决方案慢373x。在过去的7年里，这种差异可能已经得到了改善，但就我个人而言，在验证它与其他解决方案的性能之前，我不会在大型表上使用这个函数。

2009年的原始帖子如下:

有很多方法可以做到这一点，它们的性能差别很大。下面是一个优化得特别好的解决方案，包括median、ROW_NUMBERs和性能。当涉及到执行期间生成的实际I/ o时，这是一个特别优的解决方案——它看起来比其他解决方案成本更高，但实际上要快得多。

该页还包含对其他解决方案和性能测试细节的讨论。请注意，如果有多行具有相同的中位数列值，则使用唯一列作为消歧器。

就像所有的数据库性能场景一样，总是尝试在真实的硬件上用真实的数据测试解决方案——你永远不知道什么时候对SQL Server优化器的更改或环境中的某个特性会使正常快速的解决方案变慢。

SELECT
   CustomerId,
   AVG(TotalDue)
FROM
(
   SELECT
      CustomerId,
      TotalDue,
      -- SalesOrderId in the ORDER BY is a disambiguator to break ties
      ROW_NUMBER() OVER (
         PARTITION BY CustomerId
         ORDER BY TotalDue ASC, SalesOrderId ASC) AS RowAsc,
      ROW_NUMBER() OVER (
         PARTITION BY CustomerId
         ORDER BY TotalDue DESC, SalesOrderId DESC) AS RowDesc
   FROM Sales.SalesOrderHeader SOH
) x
WHERE
   RowAsc IN (RowDesc, RowDesc - 1, RowDesc + 1)
GROUP BY CustomerId
ORDER BY CustomerId;

关于你的问题，杰夫·阿特伍德已经给出了简单有效的解决方案。但是，如果您正在寻找一些计算中位数的替代方法，下面的SQL代码将帮助您。

create table employees(salary int); insert into employees values(8); insert into employees values(23); insert into employees values(45); insert into employees values(123); insert into employees values(93); insert into employees values(2342); insert into employees values(2238); select * from employees; declare @odd_even int; declare @cnt int; declare @middle_no int; set @cnt=(select count(*) from employees); set @middle_no=(@cnt/2)+1; select @odd_even=case when (@cnt%2=0) THEN -1 ELse 0 END ; select AVG(tbl.salary) from (select salary,ROW_NUMBER() over (order by salary) as rno from employees group by salary) tbl where tbl.rno=@middle_no or tbl.rno=@middle_no+@odd_even;

如果你想在MySQL中计算中位数，这个github链接会很有用。

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