I successfully joined tables in another part of my code so I followed the same pattern but this time it is not returning any result and I don't understand why.
Here are the tables I am trying to join :
- roles
| id | type |
+----+--------+
| 1 | admin |
| 2 | author |
| 3 | member |
+----+--------+
- users
+----+----------+-------+----------+-------------------+--------+
| id | username | email | password | role_id | status |
+----+----------+-------+----------+-------------------+--------+
| | | | | 3 (default value) | |
+----+----------+-------+----------+-------------------+--------+
And here is the request :
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;
}
SQL should be
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
I've omitted a few of the columns you've defined in the sql you have as I simply don't see the point in them. Not only that but they may seem ambiguous.
For example:
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
Since both 2 tables have id column, it is required to specify which table is referring by <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;
}
Try like this
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;
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.