Question

我成功地将表连接到了代码的另一部分，所以我遵循相同的模式，但是这次它不返回任何结果，我也不明白为什么。

这是我要加入的表：

-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;
}

Answer 1

SQL应该是

SELECT  
    u.id, 
    r.id AS roleid, 
    u.type,
    r.username
FROM users u
    LEFT JOIN roles r ON r.id = u.role_id
ORDER BY u.id ASC

我已经省略了在sql中定义的一些列，因为我根本看不到它们的要点。不仅如此，而且它们似乎模棱两可。

例如：

SELECT 
    users.id,             // User ID
    roles.id AS roleid,   // Role ID
    type,                 // User type
    role_id,              // Role ID (We already have it, no need for it again)
    id,                   // What ID? This is ambiguous and will fail 
    username              // User Username

Answer 2

由于两个表都具有id列，因此需要指定<table_name>.id所引用的表

public function get_users()
{
    $users_list = $this->dbh->query('SELECT users.id AS userid, roles.id AS roleid, type, role_id, username FROM users LEFT JOIN roles ON users.role_id = roles.id ORDER BY users.id ASC');
    return $users_list;
}

Answer 3

尝试这样

public function get_users()
    {
        $users_list = $this->dbh->query('SELECT u.id AS userid, r.id AS roleid, r.type, u.role_id, u.username FROM users as u LEFT JOIN roles as r ON u.role_id = r.id ORDER BY u.id ASC');
        return $users_list;
    }

联接表不返回任何结果

3 个答案: