Hi,
this is obvious and is not related to the problem. I meant something
completely different - launch an action when PHP timeouts/destroys
particular session. Regardless of whether the user accesses the page or not.
(E.g. for the need to close some records in database, but that's not
important)
Tomas
> You could simply validate the user with
>
> session_start();
> $tbaged = false;
> if (isset($_SESSION['user_id'])) {
> $user = new User($_SESSION['user_id']);
> $tbaged = true;
> else {
> $login = $_REQUEST['screename'];
> $password = $_REQUEST['pword'];
> $login = clean($login);
> $password = clean($pword);
> $user = checkLogin($login,$pword);
> if (!is_null($user)) {
> $tbaged = true;
> }
> }
>
> if (!$tbaged) {
> session_destroy();
> die("You are not logged in.");
> }
>
>
>
>
> -----Original Message-----
> From: Tomas Telensky [mailto:tomas@xxxxxxxxx]
> Sent: Monday, November 12, 2007 12:20 PM
> To: php-general@xxxxxxxxxxxxx
> Subject: Trigger an action on session timeout - feature request?
>
>
> Hi,
>
> Is there any possibility to trigger an action when the session is inactive
> for some time? I need to log users' login and logout, and so I need to know
> about logouts caused by timeout. Neither there seems to be a possibility
> of a workaround like walking through all my sessions for timeouted ones
> and destroy them myself.
>
> I have searched through the PHP doc and didn't found anything. So probably
> this is a feature request. Where should I post it? PHP's bug reporting
> system,
> unlike many common bug reporting systems, doesn't seem to distinguish
> between bugs and feature requests.
>
> I think this concept of being possible to define something like
> custom "session destructor" is obvious and useful enough to be worth
> implementing to PHP.
>
> Thanks for help,
>
> Tomas
>
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