Conversion of median calculation from MySQL to SQL Server

Question

I'm trying to convert this median calculation in MySQL to SQL Server.

http://danielsetzermann.com/howto/how-to-calculate-the-median-per-group-with-mysql/

Here is the code:

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

This is as far as I got:

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

Sadly I don't know how to get past this error:

A SELECT statement that assigns a value to a variable must not be combined with data-retrieval operations.

I know there are a few other ways to calculate median that are SQL Server specific, but not knowing how to make this conversion is eating at me.

Answer 1

Why would you convert that code when you can do something like this?

select distinct gender, percentile_cont(0.5) over (partition by gender order by height)
from heights;

SQL Server has many other approaches for such a calculation.