Send Session variable from Ajax to PHP

Question

I have 4 buttons on a page that each have their own button. When the button is clicked the page does not refresh nor does it open a new page.

When a button is pressed I want to send the ID of that button via Ajax to a php file in the form of a session. Each of the 4 buttons have a different ID (1,2,3,4). If a user clicks one button and then a second button etc, I want to add each of the clicked button Id's into this session.

This is what I have so far:

Jquery:

     $(document).ready(function() {  
     $('a[name=add]').click(function(e) {
     session = $(this).attr('id')

     $.post('sessions.php', {session:session}
    );              
     });
     });

So that simply checks if a button with the name of "add" has been clicked (all 4 of the buttons have that name), and grabs the buttons ID and sends it to sessions.php

PHP:

    <?php
    session_start();
    $_SESSION['ids'] = $_POST['session'];  
    ?>

Simply creates the session.

The problem I am having is that the PHP session is getting the ID of the first button clicked and that is all. If another button is clicked Ajax does post the button ID but the session only keeps the ID of the first button clicked.

How can I add the ID of the next buttons clicked to the session.

Example: If button 1 and 3 are clicked, I need the session to have ID 1 and 3.

Igor Parra · Accepted Answer · 2013-01-16 02:29:46Z

3

session_start();

// prepare $_SESSION['ids']
if ( ! is_array($_SESSION['ids']))
{
    $_SESSION['ids'] = array();
}

// fill $_SESSION['ids'] only with new $_POST['session'] posted
if ( ! in_array($_POST['session'], $_SESSION['ids']))
{
    $_SESSION['ids'][] = $_POST['session'];
}

edited Jan 16, 2013 at 2:29

answered Jan 16, 2013 at 2:24

Igor Parra

10.4k10 gold badges74 silver badges104 bronze badges

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

6 Comments

JPDP Over a year ago

How do I then get each individual value from the array so that I can search it in a MySQL database?

Igor Parra Over a year ago

foreach ($_SESSION['ids'] as $value) { /* use $value to search in DB */ }

JPDP Over a year ago

Thank you, however that only works if only 1 button is clicked (1 result in the array) if 2 buttons for example are clicked (2 id's stored in the session array) it does not work.

Igor Parra Over a year ago

You must have problems with basic php. Take a deep look to foreach If you use my code as stated it must work... I recommend you firebug to debug AJAX posts...

JPDP Over a year ago

Got it working, was just due to the ID's displaying as a single number, eg; 123, rather than 1, 2 ,3. Just did a quick implode to comma seperate them.

|

Jirilmon · Accepted Answer · 2013-01-16 02:27:43Z

2

This is because Jquery treats name or id as unique. Thats why your javascript function reads only first button.

Solution:

Use a common class name for the 4 buttons. (<a class="sessionclass otherclasses" href="#">1</a>)

Then change your jquery code:

 $(document).ready(function() {  
     $('a.sessionclass').click(function(e) {
         session = $(this).attr('id');    
         $.post('sessions.php', {session:session}
         );              
    });
});

Now your php session should set appropriate button value on click.

And I agree with @Lim H. Change your php code too like he suggested.

edited Jan 16, 2013 at 2:27

answered Jan 16, 2013 at 2:21

Jirilmon

1,9441 gold badge12 silver badges13 bronze badges

7 Comments

Lim H. Over a year ago

yeah this is a problem too. So OP needs to combine what OMG suggests and my solution.

ROY Finley Over a year ago

OMG has the correct answer and for storing the session, you do not need an extra php script. just use $.sessionStorage( 'foo', {data:'bar'} ); in your jQuery script...+1

wakooka Over a year ago

Even though your solution is nicer than the initial code, this is not true jQuery doesn't treat name as unique : jsfiddle.net/gZZCs

Igor Parra Over a year ago

@roy-finley sessionStorage is HTML5 functionality. Not good for compatibility. @OMG: $('a[name=add]') is correct as OP use $(this) to get their id.

Igor Parra Over a year ago

@lim-h You can have the same element's name in several forms in the same page. Totally legal... $('a[name=add]') is right :)

|

kupendra · Accepted Answer · 2015-04-23 07:06:59Z

1

<?php
    session_start();
    $id = $_POST['session']);
    if(!isset($_SESSION['ids'])) {
        $_SESSION['ids']= array();         
    }

    $_SESSION['ids'][$id] = $_POST['session']';
?>

Every time the AJAX request is made, your session id got reinitialized, hence in the end only one value remains. So you may want to create an array for session ids and every new id value get inserted into this array.

edited Apr 23, 2015 at 7:06

kupendra

1,0031 gold badge16 silver badges37 bronze badges

answered Jan 16, 2013 at 2:17

Lim H.

10.1k9 gold badges51 silver badges76 bronze badges

Collectives™ on Stack Overflow

Send Session variable from Ajax to PHP

3 Answers 3

6 Comments

7 Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

6 Comments

7 Comments

Comments

Linked

Related