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..   




error_reporting (E_ALL ^ E_NOTICE);

$userid = $_SESSION['userid'];

$username = $_SESSION['username'];



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