上一年中每个用户每月的平均帖子数-MYSQL

Question

users table:

+------------+---------+---------------------+
|    id      |  name   |created_at           |
+------------+---------+---------------------+
| 1          | AAA     | 2013-01-10 01:30:00 |             
| 2          | BBB     | 2013-02-14 01:30:00 |
| 3          | CCC     | 2013-03-29 01:30:00 |
| 4          | DDD     | 2013-04-30 01:30:00 |
| 5          | EEE     | 2013-05-10 01:30:00 |
| 6          | FFF     | 2013-06-10 01:30:00 |
| 7          | GGG     | 2013-07-10 01:30:00 |
| 8          | IIS     | 2013-08-10 01:30:00 |
| 9          | IIE     | 2013-09-10 01:30:00 |
| 10         | IIF     | 2013-10-10 01:30:00 |
| 11         | IIG     | 2013-11-10 01:30:00 |
| 12         | IIH     | 2013-12-18 01:30:00 |
| 13         | IHH     | 2013-10-18 01:30:00 |
| 14         | AHH     | 2013-02-18 01:30:00 |
| 15         | AEH     | 2013-02-18 01:30:00 |
+------------+---------+---------------------+

posts table:

+------------+-------------------+---------------------+
|    id      |  user_id          | created_at          |
| 1          | 1                 | 2013-02-10 01:30:00 |
| 2          | 2                 | 2013-03-10 01:30:00 |
| 3          | 2                 | 2013-03-10 01:30:00 |
| 4          | 7                 | 2013-04-10 01:30:00 |
| 5          | 1                 | 2013-05-10 01:30:00 |
| 6          | 8                 | 2013-05-10 01:30:00 |
| 7          | 3                 | 2013-06-10 01:30:00 |
| 8          | 6                 | 2013-07-10 01:30:00 |
| 9          | 7                 | 2013-08-10 01:30:00 |
| 10         | 9                 | 2013-09-10 01:30:00 |
| 11         | 11                | 2013-10-10 01:30:00 |
| 12         | 4                 | 2013-11-10 01:30:00 |
+------------+-------------------+---------------------+

From above tables, how i can find out the average posts per user by month in previous year. Please note: There are no posts in January and December. So, those months average would be 0.

Temp tables(Does not exist, this is only for example) : Total posts and cumulative users per month:

Monthly total_users (cumulative) in Previous year:

+--------------+-------------------+
|    month     |  total_users(cumulative)      
| Jan          | 1                 |
| Feb          | 4                 |
| Mar          | 5                 |
| Apr          | 6                 |
| May          | 7                 |
| Jun          | 8                 |
| Jul          | 9                 |
| Aug          | 10                |
| Sep          | 11                |
| Oct          | 13                |
| Nov          | 14                |
| Dec          | 15                |
+--------------+-------------------+

Monthly total_posts (Previous year):

+--------------+-------------------+
|    month     |  total_posts      |
| Jan          | 0                 |
| Feb          | 1                 |
| Mar          | 2                 |
| Apr          | 1                 |
| May          | 2                 |
| Jun          | 1                 |
| Jul          | 1                 |
| Aug          | 1                 |
| Sep          | 1                 |
| Oct          | 1                 |
| Nov          | 1                 |
| Dec          | 0                 |
+--------------+-------------------+

Equation:

average_posts_per_user:

total_posts_per_month / total_users_per_month (cumulative)

Expected result:

+--------------+-------------------+
|    month     |  avg_posts        |
| Jan          | 0                 |
| Feb          | 0.25              |
| Mar          | 0.4               |
| Apr          | 0.16              |
| May          | 0.28              |
| Jun          | 0.12              |
| Jul          | 0.11              |
| Aug          | 0.1               |
| Sep          | 0.09              |
| Oct          | 0.8               |
| Nov          | 0.7               |
| Dec          | 0                 |
+--------------+-------------------+

Thanks in advanced.

Answer 1

If you don't need rows with average = 0

SELECT
    SUBSTRING(created_at,1,7) as `month`,
    (COUNT(DISTINCT id) / COUNT(DISTINCT user_id)) as `average`
FROM
    posts 
WHERE
    YEAR(created_at) = '2013'
GROUP BY
    SUBSTRING(created_at,1,7)

And if you need rows with average = 0

SELECT
    tmonths.`month`,
    (COUNT(DISTINCT posts.id) / COUNT(DISTINCT posts.user_id)) as `average`
FROM
    (
        SELECT '2013-01' as `month`
        UNION SELECT '2013-02' as `month`
        UNION SELECT '2013-03' as `month`
        UNION SELECT '2013-04' as `month`
        UNION SELECT '2013-05' as `month`
        UNION SELECT '2013-06' as `month`
        UNION SELECT '2013-07' as `month`
        UNION SELECT '2013-08' as `month`
        UNION SELECT '2013-09' as `month`
        UNION SELECT '2013-10' as `month`
        UNION SELECT '2013-11' as `month`
        UNION SELECT '2013-12' as `month`
    ) tmonths
LEFT JOIN
    posts ON SUBSTRING(posts.created_at,1,7) = tmonths.`month`
GROUP BY
    tmonths.`month`

Answer 2

What about

SELECT
  userqry.`month`,
  total_users,
  IFNULL(month_posts,0) AS total_posts,
  IFNULL(month_posts,0)/total_users AS avg_posts
FROM
  (SELECT
    `month`,
    @usernum:=@usernum+new_users AS total_users
  FROM (
    SELECT
      DATE_FORMAT(created_at,'%M') AS `month`,
      COUNT(id) AS new_users
    FROM users
    GROUP BY 
      MONTH(created_at)
  ) AS baseview,
  (SELECT @usernum:=0) AS usernuminit
  ) AS userqry
LEFT JOIN
  (SELECT
    DATE_FORMAT(created_at,'%M') AS `month`,
    COUNT(id) AS month_posts
  FROM posts
  GROUP BY 
    MONTH(created_at)
  ) AS postqry
ON userqry.`month`=postqry.`month`

SQLfiddle

EDIT

For the year comparison either use:

SELECT
  2013 AS reportyear,
  view2013.*
FROM (
  -- query from above with "WHERE created_at BETWEEN '2013-01-01' AND '2013-12-31'" in both inner queries
) AS view2013
UNION
SELECT
  2014 AS reportyear,
  view2014.*
FROM (
  -- query from above with "WHERE created_at BETWEEN '2014-01-01' AND '2014-12-31'" in both inner queries
) AS view2015

or (recomended) run the above query multiple times from your application, everytime with a different year.