FOSUserBundle: embedding the login form and choosing its template

Question

i want to insert the login form of FOSUserBundle in my template like this:

<div id="sidebar">
     {% render "FOSUserBundle::Security::login" %}
</div>

but not to render the template that is been calling in the code of loginAction() originally.

I have thought I'd find useful the possibility of passing the template I want to render as a parameter as 'max' in this example:

<div id="sidebar">
    {% render "AcmeArticleBundle:Article:recentArticles" with {'max':
3} %}
</div>

Is that possible in symfony2? If not..

should I create another action for my bundle with the same code inside of loginAction? or should I modify the original loginAction code and write control structures?

if(currentPage == 'home')
 renderResponse('template1')
else
renderResponse('template2')

Have a look here : https://github.com/FriendsOfSymfony/FOSUserBundle/issues/399 — greg0ire, Nov 08 '11 at 09:07

David Morales · Accepted Answer · 2013-06-27T22:19:06.237

36

You can create a partial to store your plain HTML form, for example this simple one:

<form action="{{ path("fos_user_security_check") }}" method="post">
  <label>Email</label>
  <input type="text" name="_username" required="required" />

  <label>Password</label>
  <input type="password" name="_password" required="required" />

  <button type="submit">Submit</button>
</form>

Then in your layout just include your brand new partial :)

If you need the csrf token (which is highly recommended), just define the provider as a Twig global variable:

twig:
    globals:
        fos_csrf_provider: "@form.csrf_provider"

Then just add the hidden input field in your form:

  <input type="hidden" name="_csrf_token" value="{{ fos_csrf_provider.generateCsrfToken('authenticate') }}" />

edited Jun 27 '13 at 22:19

answered Jun 21 '13 at 11:23

David Morales

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6

@David,How I can implement the same form showing errors there? Thanks!! – jcarlosweb Feb 25 '14 at 12:24
4

This message: > The Symfony\Component\Form\Extension\Csrf\CsrfProvider\CsrfProviderAdapter class is deprecated since version 2.4 and will be removed in version 3.0. Use the Symfony\Component\Security\Csrf\CsrfTokenManager class instead. Made me look for a fix. Turns out you don't need the global anymore. Just add this line in your form: – Mondane Sep 29 '15 at 18:03
I did that but i got : The service twig has a dependency on a non-existent service "form.csrf_provider". Any idea about what's happening ? – Charaf Dec 01 '16 at 03:08

score 10 · Answer 2 · edited Oct 15 '13 at 01:31

10

You have to specify that you are calling a controller

{{ render(controller('FOSUserBundle:Security:login')) }}

edited Oct 15 '13 at 01:31

Taryn

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answered Oct 15 '13 at 01:06

Tom Tom

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score 1 · Answer 3 · answered Jan 14 '15 at 12:59

Since the part to include the error was not answered here and I had to dig into the FOSUserBundle Code, let me also add to this post how to add the error messages to the loginPartial.html.twig form :

Have a look at the file /vendor/friendsofsymfony/user-bundle/Controller/SecurityController.php

This controller has an loginAction which builds the error variable. All you need to do in the controller you use to render the loginPartial twig is to add to this controller the following code.

$session = $request->getSession();
    // get the error if any (works with forward and redirect -- see below)
    if ($request->attributes->has(SecurityContextInterface::AUTHENTICATION_ERROR)) {
        $error = $request->attributes->get(SecurityContextInterface::AUTHENTICATION_ERROR);
    } elseif (null !== $session && $session->has(SecurityContextInterface::AUTHENTICATION_ERROR)) {
        $error = $session->get(SecurityContextInterface::AUTHENTICATION_ERROR);
        $session->remove(SecurityContextInterface::AUTHENTICATION_ERROR);
    } else {
        $error = null;
    }

    return $this->render('YourDemoBundle:Default:index.html.twig', ['error' => $error]);

In this case you are just using the same code in your controller to generate any login error messages i.e password incorrect etc.

Now in the twig which shows the form just add the following : {% include 'YourDemoBundle:Security:loginPartial.html.twig' %}

And in the loginPartial.html.twig file put in this :

{% block fos_user_content %}
{% if error %}
    <div>{{ error.messageKey|trans(error.messageData, 'security') }}</div>
{% endif %}`Let the rest of the form be as is....and then endblock ofcourse..{% endblock fos_user_content %}`

That should do it

score 0 · Answer 4 · answered Feb 25 '12 at 00:11

0

is a single "colon" between each part

{% render "FOSUserBundle:Security:login" %}

answered Feb 25 '12 at 00:11

ejosvp

36
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5

FOSUserBundle: embedding the login form and choosing its template

4 Answers4

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