(Wordpress / Ajax) ReferenceError: Can't find variable: ajaxobject

Question

I'm working with Wordpress + Ajax and even using the proper hooks I get the error "ReferenceError: Can't find variable: ajaxobject". Of course there is some problem with my ajaxurl but I don't understand where since it seems well done to me. Can you help me?

In my functions.php

add_action( 'wp_enqueue_scripts', 'add_frontend_ajax_javascript_file', 11, 2 );
function add_frontend_ajax_javascript_file()
{
   wp_localize_script( 'ajax-script', 'ajaxobject', array( 'ajaxurl' => admin_url( 'admin-ajax.php' ) ) );
}

my jQuery/AJAX file

var itemtitle = $('#itemtitle').val();
    var itemdescription = $('#itemdescription').val();
    jQuery.ajax({
        method: 'post',
        url : ajaxobject.ajaxurl, //Why???
        dataType: "json",
        data: {
            'action':'update_portfolio_function',
            'pid' : id,
            'itemtitle' : itemtitle,
            'itemdescription' : itemdescription,
        },
        success:function(data) {
            // This outputs the result of the ajax request
            alert("Coooool");
        },
        error: function(errorThrown){
            console.log(errorThrown);
        }

    });

of course the update_portfolio_function looks like

add_action('wp_ajax_update_portfolio', 'update_portfolio_function' );
function update_portfolio_function(){
    $id = $_REQUEST['pid'];
    $title = $_REQUEST['itemtitle'];
    $description = $_REQUEST['itemdescription'];
    $attachment = array(
        'ID' => $id,
        'post_title' => $title,
        'post_content' => $description
    );
    // now update main post body
    wp_update_post( $attachment );
    die();
}

Should I use init or no_priv?

y you need ajaxobject, simply use ajaxurl

mujuonly
– mujuonly

2016-07-21 11:07:18 +00:00
Commented Jul 21, 2016 at 11:07 — mujuonly
– mujuonly, Commented Jul 21, 2016 at 11:07
@MujeebuRahman That's great, many thanks! You made my day.

middlelady
– middlelady

2016-07-21 11:22:46 +00:00
Commented Jul 21, 2016 at 11:22 — middlelady
– middlelady, Commented Jul 21, 2016 at 11:22

mujuonly · Accepted Answer · 2016-07-21 16:19:56Z

1

You just need to use ajaxurl instead of ajaxobject.ajaxurl

like below

jQuery.ajax({
        method: 'post',
        url : ajaxurl,
        dataType: "json",

edited Jul 21, 2016 at 16:19

answered Jul 21, 2016 at 16:04

mujuonly

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