虽然Justin grant的解决方案看起来很可靠，但我发现当您在给定的分区键中有许多重复值时，ASC重复值的行号最终会不按顺序排列，因此它们不能正确对齐。

以下是我的研究结果的一个片段:

KEY VALUE ROWA ROWD  

13  2     22   182
13  1     6    183
13  1     7    184
13  1     8    185
13  1     9    186
13  1     10   187
13  1     11   188
13  1     12   189
13  0     1    190
13  0     2    191
13  0     3    192
13  0     4    193
13  0     5    194

我使用Justin的代码作为这个解决方案的基础。尽管考虑到使用多个派生表效率不高，但它确实解决了我遇到的行排序问题。任何改进都会受到欢迎，因为我在T-SQL方面不是那么有经验。

SELECT PKEY, cast(AVG(VALUE)as decimal(5,2)) as MEDIANVALUE
FROM
(
  SELECT PKEY,VALUE,ROWA,ROWD,
  'FLAG' = (CASE WHEN ROWA IN (ROWD,ROWD-1,ROWD+1) THEN 1 ELSE 0 END)
  FROM
  (
    SELECT
    PKEY,
    cast(VALUE as decimal(5,2)) as VALUE,
    ROWA,
    ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY ROWA DESC) as ROWD 

    FROM
    (
      SELECT
      PKEY, 
      VALUE,
      ROW_NUMBER() OVER (PARTITION BY PKEY ORDER BY VALUE ASC,PKEY ASC ) as ROWA 
      FROM [MTEST]
    )T1
  )T2
)T3
WHERE FLAG = '1'
GROUP BY PKEY
ORDER BY PKEY

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;

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

以下是我的解决方案:

with tempa as

 (

    select value,row_number() over (order by value) as Rn,/* Assigning a 
                                                           row_number */
           count(value) over () as Cnt /*Taking total count of the values */
    from numbers
    where value is not null /* Excluding the null values */
 ),

tempb as

  (

    /* Since we don't know whether the number of rows is odd or even, we shall 
     consider both the scenarios */

    select round(cnt/2) as Ref from tempa where mod(cnt,2)=1
    union all
    select round(cnt/2) a Ref from tempa where mod(cnt,2)=0
     union all
    select round(cnt/2) + 1 as Ref from tempa where mod(cnt,2)=0
   )
  select avg(value) as Median_Value

  from tempa where rn in

    ( select Ref from tempb);

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)。

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)