CakePHP Development

How to debug Ajax calls?

• Debugging PHP, JSON and JavaScripts Application using PHP IDE
• PHP, JSON and JavaScript in Web2.0 Applications

• Place prototype.js file in app/webroot/js folder - you can get it from www.prototypejs.org
• To invoke Ajax helper add

  var $helpers = array('Html', 'Javascript', 'Ajax');

  to posts_controller.php in PostsController class
• Change index.thtml:
  • Add javascript links:

    <?php print $html->charsetTag('UTF-8'); 
    print $javascript-link('prototype');?>

  • Replace

    <?php echo $html->link( 
        'Delete', 
        "/posts/delete/{$post['Post']['id']}", 
        null, 
        'Are you sure?' 
    );

    With ajax link version:

    <?php echo $ajax->link( 
        'Delete', 
        "/posts/delete/{$post['Post']['id']}", 
        null, 
        'Are you sure?' 
    );

  Your code is now Ajax enabled and delete post action link will actually result in Ajax call to the controller. If you place DebugBreak() in delete() method of PostsController class, you will see how Ajax debugging works:

• Run index.php in external browser to avoid the deadlock:

• Click on delete now and see debugging session start: