在我的解决方案表中是一个只有分数列的学生表，我正在计算分数的中位数，这个解决方案是基于SQL server 2019的

with total_c as ( --Total_c CTE counts total number of rows in a table
    select count(*) as n from student
),
even as ( --Even CTE extract two middle rows if the number of rows are even
    select marks from student 
    order by marks 
    offset (select n from total_c)/2 -1 rows
    fetch next 2 rows only
),
odd as ( --Odd CTE extract middle row if the number of rows are odd
    select marks from student 
    order by marks 
    offset (select n + 1 from total_c)/2 -1 rows
    fetch next 1 rows only
    )
--Case statement helps to select odd or even CTE based on number of rows
select                                                        
case when n%2 = 0 then (select avg(cast(marks as float)) from even)
    else (select marks from odd)
end as med_marks
from total_c

对于像我这样正在学习基础知识的新手来说，我个人觉得这个例子更容易理解，因为它更容易理解到底发生了什么以及中值来自哪里……

select
 ( max(a.[Value1]) + min(a.[Value1]) ) / 2 as [Median Value1]
,( max(a.[Value2]) + min(a.[Value2]) ) / 2 as [Median Value2]

from (select
    datediff(dd,startdate,enddate) as [Value1]
    ,xxxxxxxxxxxxxx as [Value2]
     from dbo.table1
     )a

不过，对上面的一些代码绝对敬畏!!

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;

下面的查询从一列中的值列表返回中位数。它不能作为聚合函数使用，也不能与聚合函数一起使用，但仍然可以在内部select中使用带有WHERE子句的子查询。

SQL Server 2005+:

SELECT TOP 1 value from
(
    SELECT TOP 50 PERCENT value 
    FROM table_name 
    ORDER BY  value
)for_median
ORDER BY value DESC

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