【发布时间】:2020-02-25 10:18:44
【问题描述】:
我成功地在我的代码的另一部分加入了表格,所以我遵循了相同的模式,但这次它没有返回任何结果,我不明白为什么。
这是我要加入的表格:
- 角色
| id | type |
+----+--------+
| 1 | admin |
| 2 | author |
| 3 | member |
+----+--------+
- 用户
+----+----------+-------+----------+-------------------+--------+
| id | username | email | password | role_id | status |
+----+----------+-------+----------+-------------------+--------+
| | | | | 3 (default value) | |
+----+----------+-------+----------+-------------------+--------+
这是请求:
public function get_users()
{
$users_list = $this->dbh->query('SELECT users.id, roles.id AS roleid, type, role_id, id, username FROM users LEFT JOIN roles ON users.role_id = roles.id ORDER BY id ASC');
return $users_list;
}
【问题讨论】:
-
试试看:
$users_list = $this->dbh->query('SELECT * FROM users AS users LEFT JOIN roles AS roles ON users.id = roles.id ORDER BY id ASC'); -
它仍然没有显示任何结果。这是我遵循
SELECT column_name(s) FROM table1 LEFT JOIN table2 ON table1.column_name = table2.column_name;的模式,但我想我打错了什么,我会再检查一遍