我在代碼的另一部分成功地加入了表格，所以我遵循了相同的模式，但這次它沒有返回任何結果，我不明白為什么。以下是我要加入的表格：- 角色| id | type |+----+--------+| 1 | admin || 2 | author || 3 | member |+----+--------+- 用戶+----+----------+-------+----------+-------------------+--------+| id | username | email | password | role_id | status |+----+----------+-------+----------+-------------------+--------+| | | | | 3 (default value) | |+----+----------+-------+----------+-------------------+--------+這是請求：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;}
3 回答
森林海
TA貢獻2011條經驗 獲得超2個贊
SQL應該是
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
我省略了您在 sql 中定義的一些列，因為我根本看不到它們的意義。不僅如此，它們可能看起來模棱兩可。
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
慕的地10843
TA貢獻1785條經驗 獲得超8個贊
像這樣試試
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;
}
尚方寶劍之說
TA貢獻1788條經驗 獲得超4個贊
由于兩個表都有 id 列，因此需要指定哪個表引用<table_name>.id
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;
}
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