通常情况下，我们不仅需要为整个表计算Median，还需要为与某个ID相关的聚合计算Median。换句话说，计算表中每个ID的中位数，其中每个ID有许多记录。(基于@gdoron编辑的解决方案:性能良好，适用于许多SQL)

SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val, 
  COUNT(*) OVER (PARTITION BY our_id) AS cnt,
  ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rnk
  FROM our_table
) AS x
WHERE rnk IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;

希望能有所帮助。

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)

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;

犹斯丁上面的例子很好。但是主键的需求应该非常清楚地说明。我曾在野外见过没有密钥的代码，结果很糟糕。

我对Percentile_Cont的抱怨是它不会从数据集中给你一个实际的值。 要从数据集中获得一个实际值的“中值”，请使用Percentile_Disc。

SELECT SalesOrderID, OrderQty,
    PERCENTILE_DISC(0.5) 
        WITHIN GROUP (ORDER BY OrderQty)
        OVER (PARTITION BY SalesOrderID) AS MedianCont
FROM Sales.SalesOrderDetail
WHERE SalesOrderID IN (43670, 43669, 43667, 43663)
ORDER BY SalesOrderID DESC

简单、快速、准确

SELECT x.Amount 
FROM   (SELECT amount, 
               Count(1) OVER (partition BY 'A')        AS TotalRows, 
               Row_number() OVER (ORDER BY Amount ASC) AS AmountOrder 
        FROM   facttransaction ft) x 
WHERE  x.AmountOrder = Round(x.TotalRows / 2.0, 0)  

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