Question

所以我到处寻找这个，但我找不到这个具体的关系。下面的查询会返回用户的所有帖子以及喜欢该帖子的人数，因为我使用了GROUP BY postid，这意味着如果结果重复但具有相同的postid，那么它们将被分组。

SELECT posts.id postid,posts.post_body,posts.post_type, ALLUSERS.USERNAME,
       likes.liker,likes.target,
       plikers.*,
       COUNT(posts.id) numberOflikes

FROM posts
INNER JOIN ALLUSERS ON(ALLUSERS.USERID=posts.FROM_userid)
LEFT JOIN likes ON(likes.target=posts.id)
LEFT JOIN(SELECT USERID pl_id FROM ALLUSERS )plikers ON(pl_id=likes.liker)
GROUP BY postid

结果是......

+--------+-----------------+------------------------+-----------+-------+--------+-------+-----------+---------------+
| postid | post_body       | post_type              | USERNAME  | liker | target | pl_id | pl_un     | numberOflikes |
+--------+-----------------+------------------------+-----------+-------+--------+-------+-----------+---------------+
|     83 | Southgate       | 20&&03 Saturday/04:05  | Superuser |  NULL | NULL   |  NULL | NULL      |             1 |
|     84 | Great post!     | 10&&03 Saturday/04:07  | Superuser |     4 | 84     |     4 | dennisrec |             7 |
|     85 | How delightful? | 10&&03 Saturday/04:07  | Superuser |    43 | 85     |    43 | zerCon    |             1 |
|     87 | Cheers...       | 10&&07 Wednesday/01:53 | Superuser |  NULL | NULL   |  NULL | NULL      |             1 |
|     88 | check this out! | 20&&07 Wednesday/03:31 | Superuser |  NULL | NULL   |  NULL | NULL      |             1 |
+--------+-----------------+------------------------+-----------+-------+--------+-------+-----------+---------------+

哪个是正确的，但这只返回分组的第一个结果。所以任务成立，有没有办法在一个查询中返回所有组的所有结果？

现在显然我可以删除GROUP BY和count(*)词组，然后get multiple duplicate results，然后过滤它们以获取帖子likers的所有详细信息，但这肯定会减慢我的服务器速度。所以我已经尝试过了。

非常感谢任何帮助。

Answer 1

如果您的模型看起来像这样

MariaDB [sandbox]> select * from posts;
+------+-----------+-----------+-------------+
| id   | post_body | post_type | from_userid |
+------+-----------+-----------+-------------+
|    1 | POST1     | NULL      |           1 |
|    2 | POST2     | NULL      |           2 |
+------+-----------+-----------+-------------+
2 rows in set (0.00 sec)

MariaDB [sandbox]> select * from likes;
+------+--------+-------+
| id   | TARGET | liker |
+------+--------+-------+
|    1 |      1 |     3 |
|    2 |      1 |     7 |
|    3 |      2 |     8 |
|    3 |      2 |     6 |
+------+--------+-------+
4 rows in set (0.00 sec)

MariaDB [sandbox]> select * from users where id < 9;
+----+----------+-----------+--------+---------------------+
| id | userName | photo     | status | ts                  |
+----+----------+-----------+--------+---------------------+
|  1 | John     | john.png  |      1 | 2016-12-08 13:14:24 |
|  2 | Jane     | jane.png  |      1 | 2016-12-08 13:14:24 |
|  3 | Ali      |           |      1 | 2016-12-08 13:14:24 |
|  6 | Bruce    | bruce.png |      1 | 2016-12-08 13:14:24 |
|  7 | Martha   |           |      1 | 2016-12-08 13:14:24 |
|  8 | Sidney   |           |      1 | 2016-12-08 13:14:24 |
+----+----------+-----------+--------+---------------------+
6 rows in set (0.00 sec)

然后正如@ 1000111建议你可以

MariaDB [sandbox]> SELECT  posts.id postid,posts.post_body,posts.post_type,POSTS.FROM_USERID
    ->  , USERS.USERNAME
    ->        ,GROUP_CONCAT(likes.liker) LIKER
    ->  ,likes.target
    ->  ,GROUP_CONCAT(plikers.pl_id) pl_id
    ->        ,GROUP_CONCAT(plikers.UNAME) pl_un
    ->  ,COUNT(posts.id) numberOflikes
    ->
    -> FROM posts
    -> INNER JOIN USERS ON USERS.ID=posts.FROM_userid
    -> LEFT JOIN likes ON likes.target=posts.id
    -> LEFT JOIN(SELECT ID pl_id, USERNAME UNAME FROM USERS )plikers ON pl_id=likes.liker
    -> GROUP BY postid;
+--------+-----------+-----------+-------------+----------+-------+--------+-------+--------------+---------------+
| postid | post_body | post_type | FROM_USERID | USERNAME | LIKER | target | pl_id | pl_un        | numberOflikes |
+--------+-----------+-----------+-------------+----------+-------+--------+-------+--------------+---------------+
|      1 | POST1     | NULL      |           1 | John     | 7,3   |      1 | 7,3   | Martha,Ali   |             2 |
|      2 | POST2     | NULL      |           2 | Jane     | 6,8   |      2 | 6,8   | Bruce,Sidney |             2 |
+--------+-----------+-----------+-------------+----------+-------+--------+-------+--------------+---------------+
2 rows in set (0.00 sec)

但你应该注意警告

MYSQL获取所有分组结果一个查询，行数

1 个答案: