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)

在UDF中，写:

 Select Top 1 medianSortColumn from Table T
  Where (Select Count(*) from Table
         Where MedianSortColumn <
           (Select Count(*) From Table) / 2)
  Order By medianSortColumn

MS SQL Server 2012(及以后版本)有PERCENTILE_DISC函数，计算排序值的特定百分比。PERCENTILE_DISC(0.5)将计算中位数- https://msdn.microsoft.com/en-us/library/hh231327.aspx

试试下面的逻辑来找出中位数:

考虑一个包含以下数字的表格: 1、1、2、3、4、5所示

中位数是2.5

with tempa as 
(
    select num,count(num) over() as Cnt,
        row_number() over (order by num) as Rnum
    from temp),
tempb as
    (
        select round(cnt/2) as ref_value
        from tempa where mod(cnt,2)<>0
        union all
        select round(cnt/2) from tempa where mod(cnt,2)=0
        union all
        select round(cnt/2+1)
        from tempa where mod(cnt,2)=0
    )
select avg(num) from tempa
where rnum in (select * from tempb);
    

这段代码有点长，但很容易理解

medii是有列val的表，它有数据集， Smedi是一个cte，它将列idx作为行号，val作为medi表中的'val'，该表是升序排序的。 这是基本的数学，如果行号是奇数，那么它的中值来自smedi。 当它是偶数时，它是中间两个值的平均值。

with smedi(idx,vals) as(
                select ROW_NUMBER() over(order by val),val from medi
                )
select (case
            when (select count(*) from medi)%2!=0 then (select vals from smedi where (((select count(*) from medi)/2))=idx)
            else (select avg(vals) from smedi where idx in ((select count(*)/2 from medi),(select (count(*)/2)+1 from medi)))
            end)

更好的是:

SELECT @Median = AVG(1.0 * val)
FROM
(
    SELECT o.val, rn = ROW_NUMBER() OVER (ORDER BY o.val), c.c
    FROM dbo.EvenRows AS o
    CROSS JOIN (SELECT c = COUNT(*) FROM dbo.EvenRows) AS c
) AS x
WHERE rn IN ((c + 1)/2, (c + 2)/2);

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