My AJAX API plugin isn't working

Question

I've created a tiny AJAX plugin to count hits on my articles and bypass caching but it just isn't working and not throwing up any errors.

Can you see what I am doing wrong here?

<?php
add_action('wp_ajax_nopriv_LogHit_callback', 'LogHit_callback');
add_action('wp_ajax_LogHit_callback', 'LogHit_callback');

function HitCount() {
  ?>
  <script type="text/javascript" >
  jQuery(document).ready(function($) {
    var data = {
        action: 'LogHit_callback',
        PostId:'<?php echo get_the_ID() ?>'
    };
    jQuery.post('http://www.test.com/wp-admin/admin-ajax.php', data, function( r ) {
alert( r );
  });
 </script>
 <?php


  function LogHit_callback() {
    global $wpdb; // this is how you get access to the database

    $postId = $_POST['PostId'];
    $hits = get_post_meta((int)$postId, 'hit-counter', true);
    $hits = $hits + 1;
    update_post_meta((int)$postId, 'hit-counter', (int)$hits);
    return $postId;
    exit; // this is required to return a proper result
  }
 }

It writes the following into my HTML where I call the HitCount function

 <script type="text/javascript" >

 jQuery(document).ready(function($) {
   var data = {
    action: 'LogHit_callback',
    PostId:'86505'
};
jQuery.ajax('http://www.lazygamer.net/wp-admin/admin-ajax.php', data, function( r )      {
     alert( r );
  });

 </script>

I know hard coding the URL to admin-ajax is bad but I am trying to rule out issues. if I put an alert inside the ajax call that is written to the html it works so I know that is being called.

But the second alert in the plugin file never gets called?

Answer 1

There is no PHP function alert, that's why your function throws an error and doesn't proceed with the rest of your code. You have to return something from the LogHit_callback function using return (or echo it with echo) and then alert it with JavaScript, that is add following code to your post function:

jQuery.post( 'http://www.test.com/wp-admin/admin-ajax.php', data, function( r ) {
    alert( r );
});

I believe that this is the correct syntax for post(). Remember that you can always debug AJAX requests with browser's developer tools. You will then know exactly what's going on.