iWareWolf_ Posted June 3, 2013 Share Posted June 3, 2013 $data = mysql_query("SELECT ip,points FROM information") or die(mysql_error()); if (!empty($_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. Quote Link to comment Share on other sites More sharing options...
Jessica Posted June 3, 2013 Share Posted June 3, 2013 Well, turn on error reporting. A blank page where there should be output indicates a fatal error. Quote Link to comment Share on other sites More sharing options...
iWareWolf_ Posted June 3, 2013 Author Share Posted June 3, 2013 I would normally get errors but this gives me a blank page. Also Im using Dreamweaver CS5.5 Quote Link to comment Share on other sites More sharing options...
mac_gyver Posted June 3, 2013 Share Posted June 3, 2013 have you intentionally echoed a test string right after that block of code so that you know that code is being executed? Â what does the 'view source' of your blank page show? Quote Link to comment Share on other sites More sharing options...
iWareWolf_ Posted June 3, 2013 Author Share Posted June 3, 2013 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. Quote Link to comment Share on other sites More sharing options...
mac_gyver Posted June 3, 2013 Share Posted June 3, 2013 assuming the php code you posted is supposed to output it's result between the <body></body> tags, the implies that your database table doesn't have any rows in it. Â is that all of the php code (less any database credentials)? Quote Link to comment Share on other sites More sharing options...
Jessica Posted June 3, 2013 Share Posted June 3, 2013 Then yes, your query is not returning anything. Did you try running the query directly in MySQL? Â Also, stop using Dreamweaver. Now. Quote Link to comment Share on other sites More sharing options...
iWareWolf_ Posted June 3, 2013 Author Share Posted June 3, 2013 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 )) { Quote Link to comment Share on other sites More sharing options...
Solution jcbones Posted June 3, 2013 Solution Share Posted June 3, 2013 Mysql can't return something it doesn't have. So of course it will stop at the data call. Add some test data, test it, then comw back here with any questions. Quote Link to comment Share on other sites More sharing options...
Jessica Posted June 3, 2013 Share Posted June 3, 2013 ... What did you think would happen? Quote Link to comment Share on other sites More sharing options...
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