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

MS SQL Server 2012(及以后版本)有PERCENTILE_DISC函数，计算排序值的特定百分比。PERCENTILE_DISC(0.5)将计算中位数- https://msdn.microsoft.com/en-us/library/hh231327.aspx

关于你的问题，杰夫·阿特伍德已经给出了简单有效的解决方案。但是，如果您正在寻找一些计算中位数的替代方法，下面的SQL代码将帮助您。

create table employees(salary int); insert into employees values(8); insert into employees values(23); insert into employees values(45); insert into employees values(123); insert into employees values(93); insert into employees values(2342); insert into employees values(2238); select * from employees; declare @odd_even int; declare @cnt int; declare @middle_no int; set @cnt=(select count(*) from employees); set @middle_no=(@cnt/2)+1; select @odd_even=case when (@cnt%2=0) THEN -1 ELse 0 END ; select AVG(tbl.salary) from (select salary,ROW_NUMBER() over (order by salary) as rno from employees group by salary) tbl where tbl.rno=@middle_no or tbl.rno=@middle_no+@odd_even;

如果你想在MySQL中计算中位数，这个github链接会很有用。

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。

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

在SQL Server 2012中，您应该使用PERCENTILE_CONT:

SELECT SalesOrderID, OrderQty,
    PERCENTILE_CONT(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

参见:http://blog.sqlauthority.com/2011/11/20/sql-server-introduction-to-percentile_cont-analytic-functions-introduced-in-sql-server-2012/