根据交叉引用表为用户选择所有角色。 PDO MySQL子查询或JOIN

Question

I tried the following code and it returned an empty array when I really expected it to return several rows. My table layout is below the code. (Notice the commented version of $connection->prepare() I tried it both ways.)

In case it is not obvious, what I am trying to do is find all the role-names assigned to a certain user if I know their user_id. (the id column from the sl_user table).

function getrolesnamesbyuserid($user_id){ //returns role names
  $connection = new PDO('mysql:host=127.0.0.1;dbname='.DBNAME, DBUSER, DBPASS);
  if(!$connection){echo '<!-- DB CONNECTION ERROR -->';}else{/* echo '<!-- DB CONNECTION CONFIRMED -->'; */}
  $prep_get_roles=$connection->prepare('SELECT sl_role.name AS role_name FROM sl_role WHERE sl_role.id IN (SELECT sl_user_roles.role_id from sl_user_roles WHERE sl_user_roles.user_id = :user_id)');
  // $prep_get_roles=$connection->prepare('SELECT r.name AS role_name FROM sl_role r WHERE r.id IN (SELECT ur.role_id from sl_user_roles ur WHERE ur.user_id = :user_id)');
  $prep_get_roles->bindParam(':user_id',$user_id);
  $prep_get_roles->execute();
  return $prep_get_roles->fetchAll(PDO::FETCH_ASSOC);
}

sl_role

+-------+-------------+
| Field | Type        |
+-------+-------------+
| id    | int(11)     |
| name  | varchar(50) |
+-------+-------------+

sl_user_roles

+---------+---------+
| Field   | Type    |
+---------+---------+
| user_id | int(11) |
| role_id | int(11) |
+---------+---------+

Just as a quick test, I was calling it like this:

var_export(getrolesnamesbyuserid($_SESSION['user_id']));

Am I heading in the right direction here? If so, how do I fix it. If not, how do I re-write this? Maybe as a join, but I don't really don't have much experience with them.

Answer 1

I've tried recreating your database structure and executing your code and can't recreate the problem, which means the error is in your data. Have you checked that your data in sl_user_roles is linking correctly, and that $_SESSION['user_id'] is actually set to something? Try inserting the following dummy rows and then replace $_SESSION['user_id'] in your var_export() line with 99999:

INSERT INTO `sl_role` VALUES (99991, 'Foo');
INSERT INTO `sl_role` VALUES (99992, 'Bar');
INSERT INTO `sl_role` VALUES (99993, 'Bin');
INSERT INTO `sl_role` VALUES (99994, 'Baz');

INSERT INTO `sl_user_roles` VALUES (99999, 99991);
INSERT INTO `sl_user_roles` VALUES (99999, 99992);
INSERT INTO `sl_user_roles` VALUES (99999, 99994);

Your query should return the roles "Foo", "Bar", and "Baz". If it does so, then the error is in your data rather than your logic. You can delete the dummy rows with the following statements:

DELETE FROM `sl_role` WHERE `id` IN (99991, 99992, 99993, 99994);
DELETE FROM `sl_user_roles` WHERE `user_id` = 99999;

Obviously, check that you haven't got any users or roles with these ids before you do so. :)

Also, I would recommend restructuring your query as follows:

SELECT `r`.`name` AS `role_name`
FROM `sl_role` AS `r`
    LEFT JOIN `sl_user_roles` AS `ur`
        ON `ur`.`role_id` = `r`.`id`
WHERE `ur`.`user_id` = :user_id

This will be much more efficient than using a subquery. Whilst lengthy, the MySQL JOIN Syntax documentation is well worth a read at some point. :)

Answer 2

I was unable to reproduce your error, create database and the php script for it, everything works fine. Sure that everything is ok with your data and your query?