Refactoring code in controller with Symfony2 - php

Question

I have this code in two methods (create and update). Each time I need to update or create a new user I need to encode the user password with the salt.
$factory = $this->get('security.encoder_factory');
$encoder = $factory->getEncoder($entity);
$password = $encoder->encodePassword($entity->getPassword(), $entity->getSalt());
$entity->setPassword($password);
To avoid code duplication what should I do?
Create a new method in controller getEncondedPassword($entity) : return $encodedPassword
Add this logic to the Form using DI injecting the $encoder as required field
Add this logic to model, and pass the $encoder in the constructor of the entity object.
Thank you!

Answer 1

If your create and edit are fairly simple and pretty much the same, you can combine it to one function which actually generates and validates the form.
Some code:
class ProductController extends Controller
{
/**
* #Route("/create", name="_product_create")
*/
public function createAction()
{
$product = new Product();
return $this->productForm($product, $this->getRequest(), 'create');
}
/**
* #Route("/edit/{product_id}", name="_product_edit_id")
*/
public function editIdAction($product_id)
{
$entity_manager = $this->getDoctrine()->getEntityManager();
$product_repository = $entity_manager->getRepository('VendorBundle:Product');
$product = $product_repository->findOneBy(
array('id' => $product_id)
);
return $this->productForm($product, $this->getRequest(), 'editId');
}
protected function productForm(Product $product, Request $request, $twig_name)
{
$form = $this->createForm(new ProductType(), $product);
if ($request->getMethod() == 'POST') {
$form->bindRequest($request);
if ($form->isValid()) {
// Do whatever we want before persisting/flushing
return $this->redirect($redirect_url);
}
}
$twig_params = array(
);
return $this->render(
'VendorBundle:Product:' . $twig_name . '.html.twig', $twig_params
);
}
}
this will render create.html.twig and editId.html.twig depending on the route.
if $product->getId() === null we are creating a new entity, else we are editing.

Answer 2

I think that the correct option is the model/entity approach.
So, I leave here the my solution:
public function hashPassword($container)
{
$factory = $container->get('security.encoder_factory');
$encoder = $factory->getEncoder($this);
$password = $encoder->encodePassword($this->getPassword(), $this->getSalt());
return $password;
}
In the controller:
//hash user password
$userEntity->setPassword($userEntity->hashPassword($this->container));

Answer 3

Right now I have improved(I at least think...) the answer to this question.
I have created an class that will receive the $encoderFactory form the DI
#services.yml
parameters:
password_encoder.class: Beubi\SignatureBundle\Handler\PasswordEncoder
services:
password_encoder:
class: %password_encoder.class%
arguments: [#security.encoder_factory]
So, I create a class that will be used in Service container:
class PasswordEncoder
{
protected $encoderFactory;
public function __construct(EncoderFactory $encoderFactory)
{
$this->encoderFactory = $encoderFactory;
}
public function encodePassword($entity){
$encoder = $this->encoderFactory->getEncoder($entity);
return $encoder->encodePassword($entity->getPassword(), $entity->getSalt());
}
}
And then in my controller:
$password = $this->get('password_encoder')->encodePassword($entity);
$entity->setPassword($password);
This way, my User object doesn't have any knowledge of $factoryEncoder or how to encode an password.
I'm expecting more comments on this question...