使用Symfony 3在同一頁上呈現多個表單

Question

我想在同一頁面上顯示登錄和註冊表單。然而，似乎我們無法在控制器的不同的兩個動作中使用不同的表單變量來呈現相同的模板。

這是我的意思;

class SecurityController extends Controller { 
/** 
* @Route("/", name="login") 
*/ 
public function loginAction(Request $request) 
{ 

    //Check if the user is already authenticated 
    if($this->container->get('security.authorization_checker')->isGranted('IS_AUTHENTICATED_FULLY')) { 
     return $this->redirect($this->generateUrl('dashboard')); 
    } 

    $authenticationUtils = $this->get('security.authentication_utils'); 

    // get the error if there is one 
    $error = $authenticationUtils->getLastAuthenticationError(); 

    // last username entered by the user 
    $lastUsername = $authenticationUtils->getLastUsername(); 

    $this->addFlash(
     'welcome', 
     'Welcome back!' 
    ); 

    return $this->render(
     '::landing.html.twig', 
     array(
      // last username entered by the user 
      'last_username' => $lastUsername, 
      'error'   => $error, 
     ) 
    ); 

    $registrationform = $this->get('form.factory')->createNamedBuilder('registrationform', UserType::class); 
} 

/** 
* @Route("/register", name="register") 
*/ 
public function registerAction(Request $request) 
{ 
    //Check authentication 
    if($this->container->get('security.authorization_checker')->isGranted('IS_AUTHENTICATED_FULLY')) { 
     return $this->redirect($this->generateUrl('dashboard')); 
    } 
    // get the authentication utils 
    $authenticationUtils = $this->get('security.authentication_utils'); 
    // get the login error if there is one 
    $error = $authenticationUtils->getLastAuthenticationError(); 
    // last username entered by the user 
    $lastUsername = $authenticationUtils->getLastUsername(); 
    // build the form 
    $user = new User(); 
    $form = $this->createForm(UserType::class, $user); 
     // handle the submit (will only happen on POST) 
     $form->handleRequest($request); 
     if ($form->isSubmitted() && $form->isValid()) { 
      $user->setRoles(array('ROLE_USER')); 
      // Encode the password (you could also do this via Doctrine listener) 
      $password = $this->get('security.password_encoder') 
       ->encodePassword($user, $user->getPlainPassword()); 
      $user->setPassword($password); 

      // save the User! 
      $em = $this->getDoctrine()->getManager(); 
      $em->persist($user); 
      $em->flush(); 

      $token = new UsernamePasswordToken($user, null, 'main', $user->getRoles()); 
      $this->get('security.token_storage')->setToken($token); 
      $this->get('session')->set('_security_main', serialize($token)); 
      // ... do any other work - like sending them an email, etc 
      // maybe set a "flash" success message for the user 

      $this->addFlash(
       'success', 
       'Please complete your profile now.' 
      ); 

      $message = \Swift_Message::newInstance() 
       ->setSubject('You have successfully signed up!') 
       ->setFrom('no-reply@kampweb.com') 
       ->setTo($user->getUsername()) 
       ->setBody($this->renderView(
        ':emails:registration.html.twig'),'text/html'); 
      $this->get('mailer')->send($message); 

      return $this->redirectToRoute('update'); 
     } else{ 
      $errors = $this->get('validator')->validate($user); 
     } 

    return $this->render(
     '::landing.html.twig', 
     array('form' => $form->createView(), 
      'last_username' => $lastUsername, 
      'error' => $error, 
      ) 
    ); 
} 

}

上面我有我的登錄和註冊動作。我想將「表單」變量傳遞給我使用登錄操作呈現的相同的樹枝模板。當我嘗試這樣做時，我收到以下異常。

Variable "form" does not exist in ::landing.html.twig at line 50 
500 Internal Server Error - Twig_Error_Runtime 

Answer 1

我會做這樣的

2枝文件 -

login.html.twig和register.html.twig - 每個文件由他本人渲染動作

現在第三個樹枝文件叫做baseLogin.html.twig

在這個文件中，我通過調用渲染兩個文件（登錄和註冊）控制器

例如

baseLogin.html.twig

<div> 
    {{ render(controller('UserBundle:Security:login')) }} 
</div> 

<div> 
    {{ render(controller('UserBundle:Registration:register')) }} 
</div>