Question

don’t know better title for this, but here’s my code.

I have class user that checks form data when it’s instanciated but I get following errors/notices:

Notice: Use of undefined constant username - assumed 'username' in C:\Users\Jinxed\Desktop\WebTrgovina\app\m\Register\User.m.php on line 7

Notice: Use of undefined constant password - assumed 'password' in C:\Users\Jinxed\Desktop\WebTrgovina\app\m\Register\User.m.php on line 7

Notice: Use of undefined constant passwordc - assumed 'passwordc' in C:\Users\Jinxed\Desktop\WebTrgovina\app\m\Register\User.m.php on line 7

... and so on for every defined variable in user class.

Here’s the user class:

class User {
    function __construct(){
        $test = 'blah';
        $username; $password; $passwordc; $name; $surname; $address;
        $this->checkInput(array(username=>20, password=>20, passwordc=>20, name=>20, surname=>40, address=>40));
    }
    //array(formName=>CharacterLimit)
    private function checkInput($fields){
        foreach($fields as $field=>$limit){
            if($_POST[$field]=='' || strlen($_POST[$field])>$limit) $this->error[] = "$field must be filled in, and it must be less than or $limit characters long.";
            else $this->{$field} = $_POST[$field];
        }
    }
}

I don’t quite understand what’s the problem, I’ve tried first just creating variables and than from the main program calling checkInput method, but I get same error.

Answer 1

You should be quoting the strings that you use as array keys:

$this->checkInput(array(
    'username'=>20,
    'password'=>20,
    'passwordc'=>20,
    'name'=>20,
    'surname'=>40,
    'address'=>40));

The documentation says:

If you use an undefined constant, PHP assumes that you mean the name
of the constant itself, just as if you called it as a string (CONSTANT
vs "CONSTANT"). An error of level E_NOTICE will be issued when this
happens. See also the manual entry on why $foo[bar] is wrong (unless
you first define() bar as a constant).