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

使用COUNT聚合， 首先可以计算有多少行，并存储在一个名为@cnt的变量中。然后 你可以计算OFFSET-FETCH过滤器的参数来指定，基于数量排序， 要跳过多少行(偏移值)和筛选多少行(获取值)。

行数 跳过是(@cnt - 1) / 2。很明显，对于奇数，这个计算是正确的，因为 首先对单个中间值减去1，然后再除以2。

这也适用于偶数计数，因为表达式中使用的除法是 整数除法;所以，当一个偶数减去1时，你得到的是一个奇数。

When dividing that odd value by 2, the fraction part of the result (.5) is truncated. The number of rows to fetch is 2 - (@cnt % 2). The idea is that when the count is odd the result of the modulo operation is 1, and you need to fetch 1 row. When the count is even the result of the modulo operation is 0, and you need to fetch 2 rows. By subtracting the 1 or 0 result of the modulo operation from 2, you get the desired 1 or 2, respectively. Finally, to compute the median quantity, take the one or two result quantities, and apply an average after converting the input integer value to a numeric one as follows:

DECLARE @cnt AS INT = (SELECT COUNT(*) FROM [Sales].[production].[stocks]);
SELECT AVG(1.0 * quantity) AS median
FROM ( SELECT quantity
FROM [Sales].[production].[stocks]
ORDER BY quantity
OFFSET (@cnt - 1) / 2 ROWS FETCH NEXT 2 - @cnt % 2 ROWS ONLY ) AS D;

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;

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);

来自主人，伊兹克·本甘!

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