Question

我正在尝试执行查询以查找不是当前用户“朋友”的所有用户。例如，如果用户以'alex'身份登录，我想显示users表中不是alex的朋友（即david）的所有用户。

如果有帮助，我正在使用PHP Codeigniter。这是我提出的查询（但它什么都不返回）：

public function getUser($username) {
    $this->db->query("SELECT username FROM users WHERE username NOT IN 
    (SELECT user1 FROM friendships WHERE user2 = '$username' 
    UNION SELECT user2 FROM friendships WHERE user1 = '$username')");
    return $query->result_array();
}

MySQL表

Users

+----+----------+
| id | username |
+----+----------+
| 1  | alex     |
+----+----------+
| 2  | james    |
+----+----------+
| 3  | david    |
+----+----------+

Friendships

+----+-------+-------+
| id | user1 | user2 |
+----+-------+-------+
| 1  | alex  | james |
+----+-------+-------+
| 2  | james | david |
+----+-------+-------+

Answer 1

您应该将用户ID存储在友谊表中而不是名称中。

试试这个：

select username
from users
where username <> '$username'
    and username not in (
        select case when user1 = '$username' then user2 else user1 end
        from friendships
        where '$username' in (user1, user2)
        )

MySQL select语句，用于排除其他两个表中的字段

1 个答案: