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

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。

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)

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);

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

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)。