Access to redirectToRoute in CustomExceptionController - php

Question

In /src/AppBundle/Controller/CustomExceptionController.php I have:
namespace AppBundle\Controller;
use Symfony\Component\Debug\Exception\FlattenException;
use Symfony\Component\HttpKernel\Log\DebugLoggerInterface;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
class CustomExceptionController extends \Symfony\Bundle\TwigBundle\Controller\ExceptionController
{
public function showAction(Request $request, FlattenException $exception, DebugLoggerInterface $logger = null)
{
return $this->redirectToRoute('custom_error'); //not working
}
}
This not working because \Symfony\Bundle\TwigBundle\Controller\ExceptionController not extends class Controller. So how can I use $this->redirectToRoute in this class?

Answer 1

redirectToRoute is part of the Controller class like you mentioned.
All you need to do is create the method yourself.
First you'll need to inject the router into your CustomExceptionController (therefore you need to define your custom controller as a service in the DI)
services:
my.custom.exception_controller:
class: CustomExceptionController
arguments: [ "#twig", "%kernel.debug%", "#router" ]
twig:
exception_controller: my.custom.exception_controller:showAction
Your custom class should look like this:
class CustomExceptionController extends \Symfony\Bundle\TwigBundle\Controller\ExceptionController
{
protected $router;
public function __construct(\Twig_Environment $twig, $debug, Router $router)
{
parent::__construct($twig, $debug);
$this->router = $router;
}
public function showAction(Request $request, FlattenException $exception, DebugLoggerInterface $logger = null)
{
}
}
After that you can implement the redirectToRoute in the CustomExceptionController just like it's done in the Controller (or just create the RedirectResponse without the helper methods directly)
/**
* Returns a RedirectResponse to the given URL.
*
* #param string $url The URL to redirect to
* #param int $status The status code to use for the Response
*
* #return RedirectResponse
*/
public function redirect($url, $status = 302)
{
return new RedirectResponse($url, $status);
}
/**
* Returns a RedirectResponse to the given route with the given parameters.
*
* #param string $route The name of the route
* #param array $parameters An array of parameters
* #param int $status The status code to use for the Response
*
* #return RedirectResponse
*/
protected function redirectToRoute($route, array $parameters = array(), $status = 302)
{
return $this->redirect($this->router->generateUrl($route, $parameters), $status);
}

Answer 2

easy way with using UrlGeneratorInterface like
getContainer()->get('router')->generate( 'custom_error', ['key' => 'some val'], 0 );
getContainer() - is own funct, see manual If you need to generate a URL from a service, type-hint the UrlGeneratorInterface service:
// src/Service/SomeService.php
use Symfony\Component\Routing\Generator\UrlGeneratorInterface;
class SomeService
{
private $router;
public function __construct(UrlGeneratorInterface $router)
{
$this->router = $router;
}
public function someMethod()
{
$url = $this->router->generate( 'custom_error', [ 'key' => 'some value' ] );
}
}