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

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

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

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)。

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!!

从员工表中得到工资的中位数

with cte as (select salary, ROW_NUMBER() over (order by salary asc) as num from employees)

select avg(salary) from cte where num in ((select (count(*)+1)/2 from employees), (select (count(*)+2)/2 from employees));