How to handle forms that are used on multiple pages? - php

Question

I have a form I need to use on multiple pages:
Controller
$emailForm = $this->get('form.factory')->createNamedBuilder('form', 'email_form')
->add('email', 'email')
->add('subject', 'text')
->add('body', 'textarea')
->getForm();
$request = $this->get('request');
if ($request->getMethod() == 'POST' && $request->request->has('email_form') ) {
$emailForm->bindRequest($request);
if ($emailForm->isValid()) {
// do stuff ...
$this->get('session')->setFlash('email_sent', "Woey, mail sent successfully!");
// Redirect on the same url to prevent double posts
return $this->redirect($this->generateUrl($this->getRequest()->attributes->get('_route')));
}
}
return $this->render('Bundle:Form:index.html.twig', array('email_form' => $emailForm->createView()));
Template
{% if app.session.getFlash('email_sent') %}
<p>{{ app.session.getFlash('email_sent') }}</p>
{% endif %}
<form action="{{ path(app.request.attributes.get('_route')) }}" method="post" {{ form_enctype(email_form) }}>
{{ form_widget(email_form) }}
<p><input type="submit" class="submit" value="Send" /></p>
</form>
It's really just standard Symfony2 form, almost like from tutorial.
I can't figure how can I efficiently use it on multiple pages (in multiple controller actions) without repeating myself (too much). So far I tried:
putting the logic into Base controller, which is parent for every controller where I want to have this form. There were 2 problems with this approach:
I couldn't figure how to redirect properly to the same page
I had to call method from parent in every action, which isn't really a problem, but I guess there has to be some more elegant way
rendering controller using embedded controllers in twig. However, I couldn't figure how to redirect properly.
So, what's the common approach to such forms?
Edit:
I'm looking for a no script solution.

Answer 1

I ended up using embedded controller on every page I needed, posting to different URL and saving cookie with referrer. Then I validate form, save cookie with results, redirect back to referrer and render results (errors, thank you message...). This seems to be the best option when you are dealing with scenarios where you have to think about disabled Javascript. You can easily disable creating of redundant cookies / flash sessions using simple conditions for cases when user is posting using AJAX.

Answer 2

The Symfony2 Forms tutorial addresses your scenario as well, see the Creating Form ClassesDocs Section in which "you'll learn how to build your form in a standalone class, which is recommended as your form becomes reusable" (Ibid.).

Answer 3

You should submit the data via an ajax post.
In the controller do your thing :p
Then render the form in the twig template, without extending layouts. (like it would be currently)
Then in any view you just replace the result: $('#formDiv').html(htmlReceivedBack);
I find it the easiest to just replace the whole html div again with the new rendered html; most likely you will show a success message or some form errors.
This way the user don't have to change pages just to send a message/feedback.