iWareWolf_
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Everything posted by iWareWolf_
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Im wondering how could I change it into something MYSQL could read
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I was wondering how I could get the ip and turn it into a value that MYSQL can log/read.
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I don't understand why this isn't doing anything.
iWareWolf_ replied to iWareWolf_'s topic in PHP Coding Help
At the moment, it returns MySQL returned an empty result set (i.e. zero rows). ( Query took 0.0004 sec ) I don't have anything on it atm, but Im planning to add. Also the code seems to stop off here. while($info = mysql_fetch_array( $data )) { -
I don't understand why this isn't doing anything.
iWareWolf_ replied to iWareWolf_'s topic in PHP Coding Help
Just this. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Wolf Empire</title> </head> <body> </body> </html> There is nothing there. I got most of the code online and placed it together. -
I don't understand why this isn't doing anything.
iWareWolf_ replied to iWareWolf_'s topic in PHP Coding Help
I would normally get errors but this gives me a blank page. Also Im using Dreamweaver CS5.5 -
$data = mysql_query("SELECT ip,points FROM information") or die(mysql_error()); if (!empty($_SERVER['HTTP_CLIENT_IP'])){ $ip=$_SERVER['HTTP_CLIENT_IP']; }elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR'])){ $ip=$_SERVER['HTTP_X_FORWARDED_FOR']; }else{ $ip=$_SERVER['REMOTE_ADDR']; } $ip = ip2long($ip); while($info = mysql_fetch_array( $data )) { if($info['ip'] != null && $info['ip'] == $ip){ Print "You have $info[points] points!"; } else { mysql_query("INSERT INTO information VALUES ('$ip','0')'"); Print "You have 0 points!"; } } This is just giving me a blank page.