关注取消关注系统

Question

我在这个网站上工作，就像关注 Twitter 一样关注和取消关注用户。下面的查询有效，然后我注意到它没有显示我不关注的用户，因为该用户已被某人关注或正在关注某人

$stmt = $mysqli->prepare("SELECT * FROM town_users LEFT JOIN user_follow_list ON user_follow_list.following_id = town_users.user_id WHERE town_users.user_id != ? AND user_follow_list.follow_id IS NULL ORDER BY RAND() LIMIT 5");
$stmt->bind_param("i", $user_id);

下面是我的table结构

town_users
user_id username
| 1    | ben
| 2    | betty
| 3    | tom
| 4    | john


user_follow_list
follow_id         follower_id         following_id
|      1     |       1         |      2
|      2     |       1         |      3

在user_follow_listBen之后是betty和tom。现在 john 关注一个用户让我们说 tom,

follow_id         follower_id         following_id
|      3     |       4         |      3

现在的问题是 john 没有出现在要关注的 ben 用户列表中。

这是为了向我未关注的用户显示，以便我可以关注他们。

Answer 1

SELECT * FROM town_users 
LEFT JOIN user_follow_list ON user_follow_list.follower_id = town_users.user_id 
WHERE town_users.user_id NOT IN (SELECT following_id FROM user_follow_list WHERE 
follower_id = 1) AND town_users.user_id != 1
GROUP BY town_users.user_id ORDER BY RAND() LIMIT 5

此查询将获取所有未被 ($user_id = 1) 关注的用户。

$stmt = $mysqli->prepare("SELECT * FROM town_users 
                          LEFT JOIN user_follow_list ON user_follow_list.follower_id = town_users.user_id 
                          WHERE town_users.user_id NOT IN (SELECT following_id FROM user_follow_list WHERE follower_id = ?) 
                          AND town_users.user_id != ?
                          GROUP BY town_users.user_id ORDER BY RAND() LIMIT 5");
$stmt->bind_param("ii", $user_id, $user_id);

这里是 SQL Fiddle: http://sqlfiddle.com/#!9/96d55/23/0

关注取消关注系统

Follow Unfollow system

php

mysql

mysqli