On Aug 12, 2013, at 4:27 AM, Clifford Shuker <clifford.shu...@ntlworld.com> 
> 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:


It might help.


tedd sperling

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