Question

我正在尝试将MySQL中的这个中值计算转换为MsSql http://danielsetzermann.com/howto/how-to-calculate-the-median-per-group-with-mysql/

以下是易用性代码

SET @row_number:=0; 
SET @median_group:='';

SELECT 
    median_group, AVG(height) AS median
FROM
    (SELECT
        @row_number:=
            CASE
            WHEN @median_group = gender THEN @row_number + 1
            ELSE 1
            END AS count_of_group,
        @median_group:=gender AS median_group,
        gender,
        height,
        (SELECT 
            COUNT(*)
        FROM heights
        WHERE a.gender = gender)
        AS total_of_group
    FROM
        (SELECT 
            gender,
            height
        FROM heights
        ORDER BY gender, height)
        AS a)
    AS b
    WHERE
        count_of_group BETWEEN total_of_group / 2.0 AND total_of_group / 2.0 + 1
GROUP BY median_group

就我而言

DECLARE @row_number integer, @median_group varchar(100)
SET @row_number = 0
SET @median_group = ''

SELECT
  @row_number = (CASE WHEN @median_group = gender THEN @row_number + 1 ELSE 1 END) AS count_of_group,
  (@median_group = gender) AS median_group,
  gender,
  height,
  (SELECT COUNT(*)
   FROM heights
   WHERE a.gender = gender)
   AS total_of_group
FROM
  (SELECT
     gender,
     height
  FROM heights)
  AS a
ORDER BY a.gender , a.height

可悲的是，我不知道如何克服这个错误 - 一个为变量赋值的SELECT语句不能与数据检索操作结合使用。

非常感谢任何帮助

我知道还有一些其他方法来计算MsSql特定的中位数，但不知道如何让这种转换对我不利。

超级感谢先进

Answer 1

为什么要在做这样的事情时转换代码呢？

<br>

SQL Server还有许多其他方法可用于此类计算。

MySql到MsSql转换（中位数）

1 个答案: