How to verify if object variable is set

Question

I have a User class which is created when a user logs in

$user = new User($userId);

Now to check if a user is logged in, I tried doing

if (isset($user)) {   // user is logged in

} else {   // user is not logged in

}

However, isset() doesn't appear to work for objects? And I've also tried is_object(). Please advise! Hopefully there is a way to do this elegantly, perhaps

if ($user->isLoggedIn()) {

}

Thanks for your time!

score 12 · Accepted Answer · answered Feb 16 '10 at 19:17

12

isset() should work, object or not. You can also use

if ((isset($user)) and ($user instanceof User))

to check whether it is set and whether it is an object of the class User.

answered Feb 16 '10 at 19:17

Pekka

442,112
142
972
1,088

score 4 · Answer 2 · edited May 23 '17 at 12:30

The problem is that

new User($userid);

will always give you a User object, even though it's constructor, which probably looks up $userid in the database, may conclude that the object doesn't exist. You can throw an exception in the constructor for invalid $userids and use a try/catch construct instead of your isset() test, or set a User->valid property in the constructor of users that do exist and check for that in your test.

See also this question for some more ideas: PHP constructor to return a NULL

score 2 · Answer 3 · answered Feb 16 '10 at 19:19

2

If you edit your User class you can use $user->isLoggedIn()

class User {

private $logged_in = false;

...

public function login($uid) {
  ... login code
  $this->logged_in = true;
}

public function isLoggedIn() {
  return $this->logged_in;
}

...
}

answered Feb 16 '10 at 19:19

thetaiko

7,816
2
33
49

This is the correct answer. If you have a user and want to check if he is logged in, it's a really bad idea to use `isset($user)` to determine if the user is logged in. – MarthyM Aug 15 '17 at 09:33

score 1 · Answer 4 · answered Feb 16 '10 at 19:20

1

isset() works with objects too. In fact it will work with anything as long as:

The variable has been declared in the current scope
The variable value is different than NULL

answered Feb 16 '10 at 19:20

AlexV

22,658
18
85
122

And how to check if the variable presented if the value is NULL? – Intacto Feb 06 '20 at 16:15
@Intacto You mean $var === NULL? – AlexV Feb 07 '20 at 14:16

How to verify if object variable is set

4 Answers4