[Starting with OOP, ran in a problem]

Oh my, thank you.. Been watching the code and tutorials for a bit too long I guess and did not spot that one lol

[Starting with OOP, ran in a problem]

Okay, so I already have a website made the old fasion way and I wanted to learn OOP so I intend to completely rework the code. Since I am a total noob at this I already came across a problem I don't know the answer. Yes I did go through the basic tutorial.

I wanted to start with my ban system and made these 2 classes to check for the user IP and make sure he aint using any crappy proxies and check if his IP is in my database.

 

class Ip {

var $ip;

function GetIp() {
   $ip = $_SERVER['REMOTE_ADDR'];

   if (!empty($_SERVER['HTTP_CLIENT_IP'])) {
       $ip = $_SERVER['HTTP_CLIENT_IP'];
   } elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])) {
       $ip = $_SERVER['HTTP_X_FORWARDED_FOR'];
   }
    return $ip;

}

}

class Ban extends Ip{

var $ip;

        function __construct() {
               $this->ip = Ip::GetIp();

		}



	function CheckBan($ip){

		$query = mysql_query("SELECT * FROM Banned");
		if(mysql_num_rows($query) > 0){
		while($row = mysql_fetch_array( $query ))
		{ 
		$deny[] = $row['IP'];
		}

		if (in_array ($ip, $deny)) {
		$Status = "banned";
		return $Status; 
		}else {
			 // this is for wild card matches
			 foreach($deny as $bannedip) {
				  if(preg_match("/$bannedip/",$ip)) {
					$Status = "banned";
					return $Status;
				    } 
		}
		}
		}

	}
}

 

If the result is a string "banned" then we redirect him to the banned.php page

 

$Ban = new Ban();
if($Ban->CheckBan($Ban) == "banned")
{$page = "banned";}

 

But I get this error

Warning: preg_match() expects parameter 2 to be string, object given in C:\wamp\www\class.php

 

I understand that it means that the $ip is not a string but an object but I dont know how to transform it into a string >.<

And I dont understand why is it defined as an object, I thought return $ip in the Ip class would return a string...