MySQL - 计算每个组中的行数

I have a SQL table user_game which contains the games that a user owns:

我有一个SQL表user_game，其中包含用户拥有的游戏：

| id | user_id | game_id |
|----|---------|---------|
| 83 | 1       | 1       |
| 84 | 1       | 2       |
| 85 | 1       | 3       |
| 86 | 2       | 2       |
| 87 | 2       | 3       |
| 88 | 2       | 4       |
| 89 | 3       | 2       |

I am trying to count the number of users which have 1 game, 2 games, 3 games.. etc.

我试图计算拥有1场比赛，2场比赛，3场比赛的用户数量等。

User 1 has 3 games, User 2 has 3 games, and User 3 has 1 game. Therefore these are the results I want to achieve:

用户1有3个游戏，用户2有3个游戏，用户3有1个游戏。因此，这些是我想要实现的结果：

| no_of_games | COUNT(no_of_games) |
|-------------|--------------------|
| 1           | 1                  |
| 2           | 0                  |
| 3           | 2                  |

COUNT(no_of_games) is the number of users that have that number of games.

COUNT（no_of_games）是具有该游戏数量的用户数。

---

I can individually get the number of users for each no_of_games with this query:

我可以使用此查询单独获取每个no_of_games的用户数：

-- Select no. of users with 1 game
SELECT no_of_games, COUNT(no_of_games)
FROM
(
  -- Select no. of games each user has
  SELECT user_id, COUNT(1) as no_of_games
  FROM user_game
  GROUP BY user_id
) as A
WHERE no_of_games = 1;

which gives the results:

给出了结果：

| no_of_games | COUNT(no_of_games) |
|-------------|--------------------|
| 1           | 1                  |

However I have to change the no_of_games = 1 to 2, 3, 4... manually and UNION them with this solution and I can't do it for ~60 cases.

但是我必须手动将no_of_games = 1更改为2,3,4 ...并使用此解决方案UNION它们，我不能做~60个案例。

Is there a simpler way to achieve this?

有没有更简单的方法来实现这一目标？

2 个解决方案

#1

Your problem is a bit tricky, because groups of games which do not appear in your data with a certain frequency (e.g. 2) will not appear in the result set just using your original table. In the query below, I use a second table called nums which simply contains the sequence 1 through 10 representing counts of number of games. By using a LEFT JOIN we can retain each game count in the final result set.

您的问题有点棘手，因为没有以特定频率出现在您的数据中的游戏组（例如2）将不会仅使用原始表格出现在结果集中。在下面的查询中，我使用了第二个名为nums的表，它只包含表示游戏数量的序列1到10。通过使用LEFT JOIN，我们可以在最终结果集中保留每个游戏计数。

SELECT t1.no_of_games,
       COALESCE(t2.no_of_games_count, 0) AS no_of_games_count
FROM nums t1
LEFT JOIN
(
    SELECT t.no_of_games, COUNT(*) AS no_of_games_count
    FROM
    (
        SELECT COUNT(*) AS no_of_games
        FROM user_game
        GROUP BY user_id
    ) t
    GROUP BY t.no_of_games
) t2
    ON t1.no_of_games = t2.no_of_games
ORDER BY t1.no_of_games

And here is the definition I used for nums:

以下是我用于nums的定义：

CREATE TABLE nums (`no_of_games` int);
INSERT INTO nums (`no_of_games`)
VALUES
    (1),(2),(3),(4),(5),(6),(7),(8),(9),(10);

Demo here:

在这里演示：

SQLFiddle

#2

You can find count of games for each user and then find count of users for each count of games.

您可以找到每个用户的游戏数量，然后查找每个游戏数量的用户数量。

select cnt no_of_games, count(*) cnt_no_of_games
from(
    select user_id, count(*) cnt
    from your_table
    group by user_id
) t group by cnt;