PHP LOGOUT

[shebbycs]

<?php 
session_start();
$_SESSION['is_logged_out'] = 1;
if($_SESSION['is_logged_out'] ==1)
{
echo("<SCRIPT LANGUAGE='JavaScript'>window.alert('You had been successfully signed out')</SCRIPT>");
session_unset();
session_destroy();
header("Location:chatframe.php");
}
?>

 

I got this error  Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\login\logout.php:6) in C:\xampp\htdocs\login\logout.php on line 9

 

 

which mistake im put?

[MasterACE14]

you can't have any output before the header() function. So in your case you can't have that echo() in there.

[Guest]

echo '<meta http-equiv="refresh" content="0;url=YOURURL">';

you can use that as an alternative if you want.

[shebbycs]

you can't have any output before the header() function. So in your case you can't have that echo() in there.

 

 

thanks now im had clear my doubts

[shebbycs]

echo '<meta http-equiv="refresh" content="0;url=YOURURL">';

you can use that as an alternative if you want.

 

 

ya it is better alternative