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); ?>
