PHP Session Class

Question

I followed the wikihow tutorial for building a secure session management system : http://www.wikihow.com/Create-a-Secure-Session-Managment-System-in-PHP-and-MySQL

and it works fine.

Now, I have a problem that session id cookie _s change (and log in information) when 2 successive ajax are called like this :

<input type="button" value="go" id="mybutton" />
<script>
        $("#mybutton").click( function() {
            $.get("ajax1.php");
            $.get("ajax2.php");
        });

</script>

where both ajax1.php and ajax2.php have just require the session class file

<?php
require('session.class.php');
$session = new session();
$session->start_session('_s', false);
?>

upon clicking the button, the session id stored in _s cookie changes to a new one. I added an alert between the two ajax like this

<input type="button" value="go" id="mybutton" />
<script>
        $("#mybutton").click( function() {
            $.get("ajax1.php");
            alert("anything");
            $.get("ajax2.php");
        });

</script>

by separating the two ajax calls, the session id didn't change.

EDIT: the actual code isn't like this ,, but rather many separate buttons each have it's ajax,, but it happens that a user clicks buttons successively before response.

<input type="button" value="go" id="mybutton" />
<script>
        $("#mybutton").click( function() {
            $.get("ajax1.php");
        });

</script>
<input type="button" value="go" id="mybutton2" />
<script>
        $("#mybutton2").click( function() {
            $.get("ajax2.php");
        });

</script>

any ideas??

Barmar · Accepted Answer · 2014-06-30 00:30:08Z

3

When you make the first AJAX call, the server starts a session and sends the session ID back to the browser in a cookie. In order for the second AJAX call to be in the same session, it has to send that cookie back to the server.

But you're not waiting for the response to the first call before you send the second call. So the cookie hasn't been received, and it can't be sent with the second call. Anything that depends on the result of an AJAX call has to be done in its callback function. So you should do:

$.get('ajax1.php', function() {
    $.get('ajax2.php');
});

answered Jun 30, 2014 at 0:30

Barmar

789k57 gold badges555 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Megamind Saiko Over a year ago

thanks .. but the actual code has only one ajax in a sequence ,, but has many buttons on the page and the users click on many buttons before waiting for the response .. how can I overcome this

Barmar Over a year ago

Display some kind of "loading..." thing when they click on a button, and don't allow them to click on anything else until the response comes back.

Megamind Saiko Over a year ago

I thought of doing this,, but that will annoy users as they have to make a lot of choices, each takes about 1 second to finish,, I thought there may be another solution

Barmar Over a year ago

There's really no other solution. If you want two calls to be in the same session, you have to wait for the first one to finish before making the second one.

Barmar Over a year ago

You could display the loader only the first time they click on any button. Once you get that reply, you know the session is set up, and all other buttons can be processed without making the user wait.

|

Collectives™ on Stack Overflow

PHP Session Class

1 Answer 1

6 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related