Zend Form with Ajax and validation

Question

Following "logical" Problem, i have an Controller with 2 Actions:

IndexAction: Display an Search Form, inside the view Script is an Div Tag to display search results.

The action validates the form:

    public function indexAction()
    {
        $searchForm = new My_Form_Search();
        $statsService =  new My_Service_Statistics();

        if ($this->getRequest()->isPost()) {
            if ($searchForm->isValid($this->getRequest()->getParams())) {
                $this->_forward('ajax-do-search');
            } else {
                //???
                // i want to display the errors
                exit();
            }
        }

        $this->view->search = $searchForm;
    }

public function ajaxDoSearchAction()
{
    $this->view->result = array();
    $searchForm = new My_Form_Search();

    if ($this->getRequest()->isPost()) {
        if ($searchForm->isValid($this->getRequest()->getParams())) {
            $query = $searchForm->getValue('search');
            $search = new My_Service_Search();
            $hits = $search->find($query);
            // more...

        }
    }
}

If valid it forwards to the Search action, and the view is rendered via Jquery.form in the defined div.

But what to do if the server side validation failed? without "exit();" the index action is displayed inside the result div.

I think the solution is simple, but to much code today :-)

I know i can prevent this with client side validation, but i thrust php :-

Tomáš Fejfar · Accepted Answer · 2011-03-14 18:46:54Z

2

Your design is bad IMO.

Branch your code using $this->_request->isXmlHttpRequest().

Then echo either whole page (no-js fallback) or only the errors/results

if ($this->_request->isXmlHttpRequest()) {
    if ($form->isValid($post)) {
        $result = array('status'=>'ok', 'data' => $model->getResults());
        $this->_json($result);
    } else {
        $result = array('status'=>'error', 'data' => $form->getErrors());
        $this->_json($result);
    }
} else {
    //like you would without ajax
}

answered Mar 14, 2011 at 18:46

Tomáš Fejfar

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3 Comments

opHasNoName Over a year ago

You are right, but now i have to use json to print the result (-:

Tomáš Fejfar Over a year ago

no you don't have to ;) you can do $this->render('my-ajax-viewscript'); as well. This is just my best practice ;)

opHasNoName Over a year ago

Hi, just changed the code yesterday exactly to your comment :-) Now i see things much clearer ! Thanks a lot!

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