On Aug 12, 2013, at 4:27 AM, Clifford Shuker <clifford.shu...@ntlworld.com> wrote: > 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); > > ?> >
Ok, but when are you populating the SESSION's? Such as: $_SESSION['userid'] = $userid; Also, have a look at this: http://sperling.com/php/authorization/log-on.php It might help. tedd _______________ tedd sperling tedd.sperl...@gmail.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
