2013-07-11 35 views
1

我會以非常直接的方式驗證用戶(FOSUserBundle,Symfony2.2)。我'用一個簡單的例子很努力,但它不工作:FOSUserBundle驗證所選用戶

... 
use FOS\UserBundle\Controller\RegistrationController as RegController; 
... 
class DefaultController extends Controller{ 
... 
public function indexAction(){ 
$route = 'first_set_profile'; 
$url = $this->container->get('router')->generate($route); 
$response = new RedirectResponse($url); 
$userManager = $this->get('fos_user.user_manager'); 
$userToLogIn = $userManager->findUserByEmail('[email protected]'); 
new RegController(authenticateUser($userToLogIn, $response)); 
... 
} 

此腳本運行,但不進行身份驗證與電子郵件[email protected] ...

感謝用戶

回答

3

這是你如何能夠例如編程驗證演示用戶:

use Symfony\Component\Security\Core\Authentication\Token\UsernamePasswordToken; 

public function demologinAction(Request $request) 
{ 
    $userManager = $this->get('fos_user.user_manager'); 
    $user = $userManager->findUserByEmail('[email protected]'); 

    if (!$user) { 
     throw $this->createNotFoundException('No demouser found!'); 
    } 

    $token = new UsernamePasswordToken($user, $user->getPassword(), 'main', $user->getRoles()); 

    $context = $this->get('security.context'); 
    $context->setToken($token); 

    $router = $this->get('router'); 
    $url = $router->generate('dashboard_show'); 

    return $this->redirect($url); 
} 

UsernamePasswordToken第三個參數必須是防火牆的名稱。

+0

謝謝帕齊, 它的工作就像一個魅力! – karpatil