how to get total user count monthwise mysql

Question

How do I get total user count in mySQL by Grouping by month

I want to list the total count of registered users grouped by month.

The difficulty about this is that I want count of all the months between a range necessarily with count as 0 if there's no data during any month.

My code

SELECT tots.*, @var := @var + tots.`count` as totalCount
  FROM (
     SELECT
        YEAR(registered) AS `year`,
        MONTH(registered) AS `month`,
        COUNT(*) AS `count`
     FROM user where id = 73
          AND registered >= '2017-01-01'
          AND registered <= '2017-05-01'
     GROUP BY `year`, `month`
  )
AS tots, (SELECT @var := 0) AS inc

Current Output

+------+-------+-------+-------------+
| Year | Month | Count | TotalCount  |
+------+-------+-------+-------------+
| 2017 | 01    | 1     | 1           |
| 2017 | 04    | 4     | 5           |
+------+-------+-------+-------------+

Preferred Output

+------+-------+-------+-------------+
| Year | Month | Count | TotalCount  |
+------+-------+-------+-------------+
| 2017 | 01    | 1     | 1           |
| 2017 | 02    | 0     | 1           |
| 2017 | 03    | 0     | 1           |
| 2017 | 04    | 1     | 2           |
| 2017 | 05    | 0     | 2           |
+------+-------+-------+-------------+

Here Count is total new user in that month and TotalCount is total user up to that month.

The NULL results ARE necessary.

How could I achieve that result in mySQL or in Laravel PHP?

Answer 1

Presumably, you have at least one user registered in each month. If so, you can use conditional aggregation:

SELECT tots.*, (@var := @var + tots.`count`) as totalCount
FROM (SELECT YEAR(registered) AS `year`, MONTH(registered) AS `month`,
             SUM( id = 73 ) AS `count`
      FROM user 
      WHERE registered >= '2017-01-01' AND registered <= '2017-05-01'
      GROUP BY `year`, `month`
     ) tots CROSS JOIN
     (SELECT @var := 0) AS params;

Otherwise you need a calendar table or a way of generating one row per month.