Return Symfony's validation error as success

Question

I have a Symfony entity with @UniqueEntity annotation. I use it so I don't get double records in my database with the same email but I don't want to return to the user that the email is already in the database (privacy reasons).

How can I use @UniqueEntity but catch the error and return to the user a success message?

Controller code:

    public function register(Request $request, EntityManagerInterface $entityManager, TranslatorInterface $translator): Response
    {
        $success = false;

        $newsletter = new Newsletter();
        $form = $this->createForm(NewsletterType::class, $newsletter);
        $form->handleRequest($request);
        if ($form->isSubmitted() && $form->isValid()) {
                $newsletter = $form->getData();

                $entityManager->persist($newsletter);
                $entityManager->flush();

                $this->addFlash('success', $translator->trans('newsletter.signed_up'));

                return $this->redirectToRoute('landing');
        }

        return $this->render('landing/landing.html.twig', [
            'newsletter_form' => $form->createView(),
            'newsletter_success' => $success,
        ]);
    }

Answer 1

Basically, two possibilities:

1. Check if inserted email is already in DB performing an explicit query (keep @UniqueConstraint annotation in order to preserve data integrity).

2. Wrap entity manager operation in try-catch block, as follows (check also this answer)

   try {
     $entityManager->persist($newsletter);
     $entityManager->flush();
   } catch (UniqueConstraintViolationException $e) {
     // you can check if the exception is raised by email field, somehow, or just "mask" with a general error
   }