Question

我有一个用户表：

| username | user_id |

和朋友表格：

| user_id | friend_id |

我发布user_id以检索某个用户的朋友。

我正在使用$query = "SELECT friend_id FROM friends WHERE user_id = ?"
我也想要这些朋友的用户名，所以我想加入这些表。

$query = 'SELECT f.friend_id,u.username FROM friends f WHERE f.user_id = ? JOIN users u ON u.user_id = f.friend_id'

但在这种情况下结果为零。我的逻辑错了吗？

Answer 1

更新您的MYSQL查询：

$query = 'SELECT f.friend_id,u.username FROM friends f WHERE f.user_id = ? JOIN users u ON u.user_id = f.friend_id'

使用此查询：

$query = 'SELECT f.friend_id, u.username FROM friends f JOIN users u ON u.user_id = f.friend_id WHERE f.user_id = ?'

无论我们在表格中使用MYSQL JOIN ... ON ，都会将此项置于 WHERE 之类的条件子句之前。

Answer 2

你的查询应该是这个

SELECT f.friend_id,u.username FROM friends f JOIN users u ON u.user_id = f.user_id WHERE f.user_id = ?

Answer 3

$query = 'SELECT f.friend_id, u.username FROM friends f JOIN users u ON u.user_id = f.friend_id where f.user_id = ?'

Answer 4

$query = "select f.`friend_id`, u.`username` 
          from `users` u 
          join `friends` f on u.user_id = f.user_id
          join `users` u2 on u2.user_id = f.user_id
          where u2.`user_id` = ?"

Answer 5

这样的事情应该有效

$query = "select f.`friend_id`, u.`username` from `friends` f
          left outer join `users` u on u.`user_id`=f.`friend_id` 
          where f.`user_id` = ?";

没有发现初始错误...

Answer 6

应该是

$query = 'SELECT f.friend_id,u.username FROM friends f WHERE f.user_id = ? ANd u.user_id = f.user_id'

SQL连接表将id转换为用户名

6 个答案: