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 htmlheadcontaining
meta info title/headbodycontaining 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