I am struggling with this error for a while now.

The plugin inserts data to the table.

Here's the code :

File : plugin/_add.php


global $wpdb;
$table_name = $wpdb->prefix . 'westpac_store';
$result = $wpdb->insert( 
                'itemModel' => $_POST['itemModel'], 
                'itemCat' => $_POST['itemCat'],
                'itemManu' => $_POST['itemManu'], 
                'itemImg' => $_POST['itemSpec'], 
                'itemSpec' => $_POST['itemDesc'], 
                'itemDesc' => $_POST['itemImg'], 
                'itemPrice' => $_POST['itemPrice'], 
                'itemPriceDiscount' => $_POST['itemDiscount'], 

return ( $result == true ? 'success' : 'error' );

File : plugin/_common.js

jQuery(document).ready(function() {

    //Add Item Submit

        var $form = jQuery(this);
        // let's select and cache all the fields
        var $inputs = $form.find("input, select, button, textarea");
        // serialize the data in the form
        var itemData = $form.serialize();

        $inputs.prop("disabled", true);

            cache: false,
            type: 'POST',
            url: store.directory + '_add.php',
            data: itemData,
            success: function(data) {
            error: function(data) {
                alert(data.error + 'error');

        $inputs.prop("disabled", false);

Facts: 1. the _add.php is not the main plugin file. 2. its an admin interface. 3. the plugin works, the data saves in the database unless there is a data type mismatch from the input and in the database.

Problem is I can't get the actual return value from _add.php, the console.log(data) always return "" even though the return in _add.php is fixed with string returns. It should be at least any of the string 'success' or 'error'.

Please help!

  • Generally just use the WordPress way of doing AJAX as described at the codex article AJAX in Plugins. – Nicolai Oct 7 '14 at 14:01
  • that doesnt solve the problem and evaded it – Roi Oct 7 '14 at 14:03
  • Never claimed that it will solve you problem. Just telling you that your life will be easier if you do it that way. – Nicolai Oct 7 '14 at 14:08
  • actually I used that documentation to produce the external ajaxing, everything works just the return thing. although I manage to figure it out. – Roi Oct 7 '14 at 14:10
  • Good for you. It still is better to do it different, but do as you like. I just wanted to give you that hint. – Nicolai Oct 7 '14 at 14:13

I solved it

instead of using

return ( $result == true ? 'success' : 'error' );


echo ( $result == true ? 'success' : 'error' );

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