Question

我有社交应用程序，比如说twitter，用户可以关注其他用户，也可以点评。

我需要获取的是用户的关注者以及每个关注者喜欢的最后2条评论的ID（如果有的话），以及使用MySQL的关注者关注者的数量。

以下是表格

表 user_follower

User_id  follower_id
1          2
2          3
1          5
1          6
1          7

表 user_likes

comment_id User_id date
41          2      some_date
42          2      some_date
41          5      some_date
42          5      some_date
43          5      some_date
43          2      some_date
43          6      some_date

我们如何在单个mysql查询中执行此操作？

到目前为止，我能够得到关注者和关注者的关注者并跟随他们。

select uf.follower_id, 
(select count(*) from  user_followers uf1 where uf1.follower_id = uf.follower_id) as following_count,
(select count(*) from  user_followers uf2 where uf2.user_id = uf.follower_id) as follower_count,
from user_followers uf 
join users u on u.id = uf.follower_id
where uf.user_id = 1

我想要的是为每个关注者获取2个最新的comment_ids，即uf.follower_id。如果在同一查询中不可能，即使使用另一个查询，我也很好，因为它会在参数中传递follower_ids，但它应该为每个传递的ID提供2条最新评论..

Answer 1

我认为这会奏效。我没有测试它，所以它可能有一些语法错误。鉴于此查询中使用的嵌套级别，我怀疑它对于非常大的数据集会表现不佳。

SELECT follower_id, num_followers, GROUP_CONCAT(comment_id)
FROM (
    SELECT t.*,
        @r := IF(@g = t.follower_id, @r+1, 1) RowNum,
        @g := t.follower_id
    FROM (select @g:=null) AS initvars
    INNER JOIN (
        SELECT followers.*, ul.comment_id
        FROM (
            SELECT
                uf1.user_id,
                uf1.follower_id,
                COUNT(uf2.follower_id) AS num_followers
            FROM user_follower uf1
            LEFT JOIN user_follower uf2
                ON uf1.follower_id = uf2.user_id
            WHERE uf1.user_id = 1
            GROUP BY uf1.user_id, uf1.follower_id
        ) AS followers
        LEFT JOIN user_likes ul
            ON followers.follower_id = ul.user_id
        ORDER BY followers.follower_id ASC, comment_id DESC
    ) AS t
) AS final
WHERE RowNum < 3
GROUP BY follower_id, num_followers;

UPDATE 以下是使用不等式连接的其他查询 -

SELECT tmp.follower_id, COUNT(uf2.follower_id) AS num_followers, tmp.comments
FROM (
    SELECT follower_id, GROUP_CONCAT(comment_id ORDER BY comment_id DESC) AS comments
    FROM (
        SELECT uf.follower_id, ul1.*
        FROM user_follower uf
        LEFT JOIN user_likes ul1
            ON uf.follower_id = ul1.user_id
        LEFT JOIN user_likes ul2
            ON uf.follower_id = ul2.user_id
            AND ul1.comment_id < ul2.comment_id
        WHERE uf.user_id = 1
        GROUP BY ul1.user_id, ul1.comment_id
        HAVING COUNT(ul2.comment_id) < 2
    ) AS tmp
    GROUP BY tmp.follower_id
) AS tmp
LEFT JOIN user_follower uf2
    ON tmp.follower_id = uf2.user_id
GROUP BY tmp.follower_id

用户的关注者和每个关注者喜欢的最后2条评论的ID（如果有的话），也是关注者关注者的数量，使用mysql

1 个答案: