我尝试了几种替代方案，但由于我的数据记录有重复的值，ROW_NUMBER版本似乎不是我的选择。这里是我使用的查询(NTILE版本):

SELECT distinct
   CustomerId,
   (
       MAX(CASE WHEN Percent50_Asc=1 THEN TotalDue END) OVER (PARTITION BY CustomerId)  +
       MIN(CASE WHEN Percent50_desc=1 THEN TotalDue END) OVER (PARTITION BY CustomerId) 
   )/2 MEDIAN
FROM
(
   SELECT
      CustomerId,
      TotalDue,
     NTILE(2) OVER (
         PARTITION BY CustomerId
         ORDER BY TotalDue ASC) AS Percent50_Asc,
     NTILE(2) OVER (
         PARTITION BY CustomerId
         ORDER BY TotalDue DESC) AS Percent50_desc
   FROM Sales.SalesOrderHeader SOH
) x
ORDER BY CustomerId;

如果你使用的是SQL 2005或更好的版本，这是一个很好的，简单的中位数计算表中的单列:

SELECT
(
 (SELECT MAX(Score) FROM
   (SELECT TOP 50 PERCENT Score FROM Posts ORDER BY Score) AS BottomHalf)
 +
 (SELECT MIN(Score) FROM
   (SELECT TOP 50 PERCENT Score FROM Posts ORDER BY Score DESC) AS TopHalf)
) / 2 AS Median

查看SQL中位数计算的其他解决方案: “用MySQL计算中位数的简单方法”(解决方案大多与供应商无关)。

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

with t1 as (select *, row_number() over(order by ordqty) as rn,
count(*) over() as rc from ord_line)
select rn,* from t1 where rn in((rc+1)/2, (rc+2)/2);

它将计算偶数和奇数的中位数。

Ord_line是一个表 Ordqty是一个列

对于连续变量/测量'table1'中的'col1'

select col1  
from
    (select top 50 percent col1, 
    ROW_NUMBER() OVER(ORDER BY col1 ASC) AS Rowa,
    ROW_NUMBER() OVER(ORDER BY col1 DESC) AS Rowd
    from table1 ) tmp
where tmp.Rowa = tmp.Rowd