PHP AJAX Session Variable Not Working

Question

I am new to using Session variables and have been struggling despite scouring the net to understand.

So basically I have a page (search.php). I load a dynamic data table based on a search form. Once the table is loaded, via AJAX I perform the following:

$.ajax({
    type:'POST',
    url:'/ITSMIS/data/asset/search.php',
    data:HardwareAsset,
    dataType: 'html',
    // When PHP / SQL Query Has Been Executed If Data Is Returned
    success:function(data){
        // Load The Data Table Results
        LoadDataTable(data);
        // Generate Hyperlinks For The Table Rows

        $('#data-table tr').click(function(){       
            var HyperlinkHardwareAssetID = $(this).data('id');
            $.ajax({
                type:'POST',
                url:'/ITSMIS/session.php',
                data: HyperlinkHardwareAssetID,
                success: function(data){
                    alert("great!");
                }
            });
            window.location = $(this).data('href');
        }); 

    }
})

As you can see on Success and on click of a table row record, the idea is I post a variable via AJAX to use later. And redirect the user to the update.php page.

I am expecting to then create a session variable in the session.php page using the POSTED AJAX variable.

session_start();
$_SESSION["HardwareAssetID"] = $_POST["HyperlinkHardwareAssetID"];

I am then expected when the update.php page loads to display the Session variable result on the page.

        $session = $_SERVER['DOCUMENT_ROOT'];
        $session .= "/ITSMIS/session.php";
        include_once($session);

        echo $_SESSION["HardwareAssetID"];

But I only get the following error:

Notice: Undefined index: HyperlinkHardwareAssetID in C:\xampp\htdocs\ITSMIS\session.php on line 3

Implying for some reason the AJAX POST on click of the table row hasnt worked. However the alert with the success is always triggered.

Any ideas???

Christian Valentin · Accepted Answer · 2017-04-23 20:38:08Z

1

To send variables to your php, you have to send an object in data property of $.ajax function, like:

$.ajax({
    data: {
        'var_name': 'value'
    }
});

and, in the PHP file:

$item = $_POST['var_name'];

So, in your code, you should use your $.ajax function this way:

$.ajax({
   type:'POST',
   url:'/ITSMIS/session.php',
   data: {
       'HyperlinkHardwareAssetID': HyperlinkHardwareAssetID
   },
   success: function(data){
       alert("great!");
   }
});

For more information, see jQuery.ajax() Documentation

answered Apr 23, 2017 at 20:38

Christian Valentin

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3 Comments

TheTechnicalPaladin Over a year ago

thanks, after you shared this info, it made sense why I needed to do this, I have done it everywhere else in my code, however this hasnt resolved my error message, any ideas?

Christian Valentin Over a year ago

Insert a print_r($_POST) in your PHP file, and in the $.ajax, insert a new data variable, like: data: {'HyperlinkHardwareAssetID': HyperlinkHardwareAssetID, 'hello': 'Hi there!'}. It seems like your HyperlinkHardwareAssetID has an undefined value, and those values doesn't appear in the $_POST variable. If you want to get the tr attribute with jQuery, use $(this).attr('id') instead of $(this).data('id'), and do the same with $(this).data('href')

TheTechnicalPaladin Over a year ago

On closer look at the code, when i try and check what "data" is in success, it says undefined....

FAROUK BLALOU · Accepted Answer · 2017-04-23 20:50:58Z

1

You need to pass the ajax data param as an object {key:value}

   $.ajax({
            type:'POST',
            url:'/ITSMIS/session.php',
            data: {HyperlinkHardwareAssetID : HyperlinkHardwareAssetID  },
            success: function(data){
                alert("great!");
            }
    });

edited Apr 23, 2017 at 20:50

answered Apr 23, 2017 at 20:35

FAROUK BLALOU

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1 Comment

TheTechnicalPaladin Over a year ago

I thanks for the advice, i noticed i was originally missing this, I have since tried this but still the same error.

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