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Peanutz's Achievements


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  1. Oh my, thank you.. Been watching the code and tutorials for a bit too long I guess and did not spot that one lol
  2. 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 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...
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