Clifford Shuker
2013-08-12 08:27:59 UTC
Hi List,
Hi have the following (below) session code at the top of each page.. The
'print_r' (development feature only) confirms that on one particular page I
do log out as the session var = (). but, on testing that page via the URL I
still get to see the page and all its contents - session var() -.. the page
has the following 'session_start, DOCTYPE Info then <html><head>containing
meta info & title</head><body>containing style/tables/content/</body></html>
// end of page. I have copied the same page without the html content (i.e.
a blank page) and I get to fully log out.. when this page is tested in the
URL my warning comes up 'you need to login to see this page' which is what I
want but, I've tried numerous avenues to reconcile my problem to no avail..
I'm a novice so any help would be appreciated..
<?php
session_start();
error_reporting (E_ALL ^ E_NOTICE);
$userid = $_SESSION['userid'];
$username = $_SESSION['username'];
print_r($_SESSION);
?>
Hi have the following (below) session code at the top of each page.. The
'print_r' (development feature only) confirms that on one particular page I
do log out as the session var = (). but, on testing that page via the URL I
still get to see the page and all its contents - session var() -.. the page
has the following 'session_start, DOCTYPE Info then <html><head>containing
meta info & title</head><body>containing style/tables/content/</body></html>
// end of page. I have copied the same page without the html content (i.e.
a blank page) and I get to fully log out.. when this page is tested in the
URL my warning comes up 'you need to login to see this page' which is what I
want but, I've tried numerous avenues to reconcile my problem to no avail..
I'm a novice so any help would be appreciated..
<?php
session_start();
error_reporting (E_ALL ^ E_NOTICE);
$userid = $_SESSION['userid'];
$username = $_SESSION['username'];
print_r($_SESSION);
?>