Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

cakephp 2 ajax form

I'm having trouble building an ajax form in cakephp 2 which obviously has changed a lot since 1.3.

I'm using the following code:

<div id="commentForm">
<div id="commentStatus"></div>
<?php
echo $this->Form->create('Comment', array('action' => 'save', 'default' => false));
echo $this->Form->input('Comment.comments_name');
echo $this->Form->input('Comment.comments_email');
echo $this->Form->input('Comment.comments_text');
echo $this->Js->submit('Save', array('update' => '#commentStatus'));
echo $this->Form->end();
?>

However the form is not submitted when pressing the button.

I will be thankful for any help!

Thanks!

like image 384
Kik Minev Avatar asked Dec 12 '22 03:12

Kik Minev


1 Answers

Try this in your view file:

<?php

    $data = $this->Js->get('#CommentSaveForm')->serializeForm(array('isForm' => true, 'inline' => true));
    $this->Js->get('#CommentSaveForm')->event(
          'submit',
          $this->Js->request(
            array('action' => 'save'),
            array(
                    'update' => '#commentStatus',
                    'data' => $data,
                    'async' => true,    
                    'dataExpression'=>true,
                    'method' => 'POST'
                )
            )
        );
    echo $this->Form->create('Comment', array('action' => 'save', 'default' => false));
    echo $this->Form->input('Comment.comments_name');
    echo $this->Form->input('Comment.comments_email');
    echo $this->Form->input('Comment.comments_text');
    echo $this->Form->end(__('Submit'));
    echo $this->Js->writeBuffer();

?>

NOTE: #CommentSaveForm is ID generated by CakePHP, If you have your own then use that

like image 73
thecodeparadox Avatar answered Dec 29 '22 06:12

thecodeparadox