doublea2k7 Posted May 1, 2008 Share Posted May 1, 2008 can anyone tell me whats wrong with this script i get this error Parse error: syntax error, unexpected $end in /home/urzone/public_html/smashlegacies.exofire.net/loginfix_check.php on line 28 <? include "config2.php"; ?> <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> </head> <body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080" alink="#ff0000"> <? $name=mysql_real_escape_string($name); if($rec=mysql_fetch_array(mysql_query("SELECT * FROM accounts WHERE name='$name'"))){ if(($rec['name']==$name)){ if(mysql_query("update accounts set loggedin='1' WHERE name='$name'")){ echo "<font face='Verdana' size='2' ><center>OK <br> Your account has been fix try to log in the game now!</font></center>";}} ?> </body> </html> im really new at php coding so please help Quote Link to comment Share on other sites More sharing options...
pocobueno1388 Posted May 1, 2008 Share Posted May 1, 2008 You didn't close a bracket, if you line your code up it will be a lot easier to spot. <? include "config2.php"; ?> <!doctype html public "-//w3c//dtd html 3.2//en"> <html> <head> </head> <body bgcolor="#ffffff" text="#000000" link="#0000ff" vlink="#800080" alink="#ff0000"> <? $name=mysql_real_escape_string($name); if ($rec=mysql_fetch_array(mysql_query("SELECT * FROM accounts WHERE name='$name'"))) { if (($rec['name']==$name)) { if (mysql_query("update accounts set loggedin='1' WHERE name='$name'")) { echo "<font face='Verdana' size='2' ><center>OK <br> Your account has been fix try to log in the game now!</font></center>"; } } } Quote Link to comment Share on other sites More sharing options...
doublea2k7 Posted May 1, 2008 Author Share Posted May 1, 2008 o ty but now i have another problem my message in echo does not display as well as the database field loggedin does not change to 1 Quote Link to comment Share on other sites More sharing options...
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