I am trying to add a simple form to allow my users to edit their profile. My problem is:
Since the entity "linked" to the form is the same as the current user object ($user === $entity, see below), if the form validation fails, then the view is rendered with the modified user object (ie. with values of the non-valid form).
Here my (classic) controller:
public function profileAction()
{
$em = $this->getDoctrine()->getEntityManager();
$user = $this->get('security.context')->getToken()->getUser();
$entity = $em->getRepository('AcmeSecurityBundle:User')->find($user->id);
// $user === $entity => true
$form = $this->createForm(new ProfileType(), $entity);
$request = $this->getRequest();
if ($request->getMethod() === 'POST')
{
$form->bindRequest($request);
if ($form->isValid()) {
$em->persist($entity);
$em->flush();
return $this->redirect($this->generateUrl('profile'));
}
}
return $this->render('AcmeSecurityBundle:User:profile.html.twig', array(
'entity' => $entity,
'form' => $form->createView(),
));
}
So I wondered how to have two distincts objects $user and $entity. I used clone() and it worked well for the view rendering part (the $user object was not modified), but it created a new record in database instead of updating the old one.
PS: I know I should use FOSUserBundle. But I would really like to understand my mistake here :)