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。

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)。

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

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

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);

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