cakePHP basic ajax form problem

Question

i am trying to do a simple form that adds a new comment using ajax to a blog post (actually this is a part of a cakePHP tutorial)

but the problem is that the submit button do nothing at all

here is the part of code that generates the form in the view.ctp file

<?php echo $ajax->form('/comments/add', 'post', array('url' => '/comments/add', 'update' => 'PostsComments', 'indicator' => 'commentSaved'));?>
<fieldset>
    <legend><?php __('Add Comment');?></legend>
        <?php  
            echo $form->hidden('Comment.post_id', array('value' => $post['Post']['id']));
            echo $form->input('Comment.name');
            echo $form->input('Comment.content');                            
        ?>
</fieldset>
<?php echo $form->end('Submit');?>

and here is the add comment action in the comments_controller.php

if (!empty($this->data)) {
    $this->Comment->create();
    if ($this->Comment->save($this->data)) {
        $comments = $this->Comment->find('all',array('conditions'=>array('post_id'=>$this->data['Comment']['post_id']),'recursive'=>-1));
        $this->set(compact('comments'));
        $this->render('add_success','ajax');
    } else {
        $this->render('add_failure','ajax');
    }
}

the problem is that the action add isnt called from the view ... when i viewed the generated html source code i found something like that

<form id="form304217296" onsubmit="event.returnValue = false; return false;" method="post" action="/php-cake/blog/comments/add"><fieldset style="display:none;">

if i removed the on submit tag manually then the action is called but the add_success.ctp is generated as a new page not as an ajax call.

so what could be the problem ?

Ahmed Kotb · Accepted Answer · 2010-02-11 11:58:50Z

3

finally i have discovered the problem

there was more than one mistake in the tutorial (chapter 8 in CakePHP from novice to professional) after fixing those errors i found another problem is that cakephp v1.2.6 is not compatible with the latest version of prototype (v1.6.1) as it gives the following error in firebug

Event.observe is not a function

so i used version 1.6.0.3 of prototype and the problem was solved.

for a list of mistakes in this chapter see this

answered Feb 11, 2010 at 11:58

Ahmed Kotb

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Comments

harpax · Accepted Answer · 2010-02-10 21:27:59Z

0

Do you include prototype and scriptaculus in your View/Layout (both are needed for the Ajax-helper)? Also there might be an interference with jquery if you use that too

answered Feb 10, 2010 at 21:27

harpax

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3 Comments

Ahmed Kotb Over a year ago

yes iam using the prototype library but i had put it in app/webroot/js not in the view/layout folder also i didnt use scriptaculus as i think it isnot needed for this simple form ? right ?

harpax Over a year ago

I am not sure about scriptaculus (but would think that you don't need that for any ajax-call). What i meant by include in your view/layout: do you have a $javascript->link('prototype'); anywhere in your view or layout ctp, so that in the source code there is a <script type="text/javascript" src="../prototype.js">?

Ahmed Kotb Over a year ago

yes i have put the line $javascript->link('prototype'); in the default .ctp found in the layout folder.

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