Ok, i've been stuck on this for some time...

I have my own module where I alter the form to add my own button that returns a message from the server via AJAX. I'm stuck on the ajax part. When I click on the button, I see the url change to ../myforms/ajax But I get the same page returned as output instead of my ajax message

Here are my code snippets.

I click on the button that calls my getAjaxReply function:

function getAjaxReply(context, settings)
console.log("this works");
$url = this.href;
            url: '/myforms/ajax',
            dataType: 'json',
    return false;

In my php file, I have

function myforms_menu()
    $items['myforms/ajax'] = array(
    'title' => t('myforms AJAX'),
    'type' => MENU_CALLBACK,
    'page callback' => 'myforms_ajaxCallback',
    'access arguments' => array('access content'),

 return $items;


//the php function function myforms_ajaxCallback() { if($_POST['getAjaxReply'] == 'testingAjax') { return drupal_json( array( 'answer' => 'success' ) ); } return false; }

  • How are you implementing myforms_ajaxCallback function? Commented Jun 19, 2012 at 12:42
  • OK updated the question with the myforms_ajaxCallback function
    – airnet
    Commented Jun 20, 2012 at 11:26
  • Just as an FYI, Drupal doesn't like camel case in PHP.
    – Chapabu
    Commented Jun 20, 2012 at 11:44
  • Have you cleared the menu cache?
    – Chapabu
    Commented Jun 20, 2012 at 11:45

2 Answers 2


you also may consider using Drupal Ajax form API. This way you don't have to write any JS code or hook_menu entries. You basically alter your form to take user input into account. Just take a look at the examples: http://drupal.org/node/752056

I recently ajaxified form by hand and after that reworked it to make use of api. And didn't regret ;)


Do not return when calling drupal_json()

function myforms_ajaxCallback(){
    if($_POST['getAjaxReply'] == 'testingAjax'){
      drupal_json(array('answer' => 'success'));
      drupal_json(array('answer' => 'failure'));

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