javascript not working with php code for session timeout checking

Question

I'm using the JavaScript below to check if the user session has not expired (with PHP) at intervals of 5 seconds. After logging out from another tab (same browser) or which the current session must have expired, still the code is bringing up the alert("Logged back in."); which is supposed to work for active sessions.

setInterval(function () {
    var checklogin = "<? if ($this->session->userdata('user_logged_in')) {echo '1';} else {echo '0';} ?>";
    if (checklogin == "1" ){
        alert("Logged back in.");
    } else {
        alert("you are already logged out!");
    } 
}, 5000);

Please what am I doing wrong ?

You need to be doing this with AJAX. The PHP here is just rendered once (on page load). — musicnothing, May 28 '14 at 20:55
sidenote, better to always use `session->userdata('user_logged_in') ? '1' : '0'; ?>` — Lawrence Cherone, May 28 '14 at 21:03
@Alex to be doing it with AJAX; how can i achieve that pls ? — getapay, May 29 '14 at 05:37
@Loz noted. i've adopted that. And can you give me the link to the place where the duplicate is? — getapay, May 29 '14 at 05:39

score 0 · Answer 1 · edited May 23 '17 at 11:52

0

The php section of your code will not change at run time. You could, for example, perform an AJAX call to your server to retrieve session information.

EDIT:

If you're using jQuery, take a look at: jQuery Ajax

If without jQuery, take a look at: How to make an ajax call without jquery?

edited May 23 '17 at 11:52

Community

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answered May 28 '14 at 21:00

grim

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how can i perform an AJAX call to the server pls? so, it will be run at interval, with the code ?? – getapay May 29 '14 at 05:34

javascript not working with php code for session timeout checking

1 Answers1