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Enfernikus · 10-05-2008, 08:05 PM

Let's say the name of your DB class is MySQLAccess

PHP Code:


			

<?php

class user
{
    private $db;
        
        public function __construct(MySQLAccess $db)
        {
            $this->db = $db;
        }
}

$db = new MySQLAccess($dbInfo);
$user = new user($db);

?>

This way the user class will NEED the object MySQLAccess to start up or it'll give off a fatal error.

10-05-2008, 08:05 PM	#2 (permalink)
Enfernikus The Addict Join Date: Jun 2008 Posts: 335 Thanks: 2	Let's say the name of your DB class is MySQLAccess PHP Code: `<?php class user { private $db; public function __construct(MySQLAccess $db) { $this->db = $db; } } $db = new MySQLAccess($dbInfo); $user = new user($db); ?>` This way the user class will NEED the object MySQLAccess to start up or it'll give off a fatal error.