在 CustomExceptionController 中访问重定向到路由

Question

在/src/AppBundle/Controller/CustomExceptionController.php 我有：

namespace AppBundleController;
use SymfonyComponentDebugExceptionFlattenException;
use SymfonyComponentHttpKernelLogDebugLoggerInterface;
use SymfonyComponentHttpFoundationRequest;
use SymfonyComponentHttpFoundationResponse;
class CustomExceptionController extends SymfonyBundleTwigBundleControllerExceptionController
{
    public function showAction(Request $request, FlattenException $exception, DebugLoggerInterface $logger = null)
    {
        return $this->redirectToRoute('custom_error'); //not working
    }
}

这不起作用，因为\Symfony\Bundle\TwigBundle\Controller\ExceptionController没有扩展类控制器。那么如何在此类中使用$this->redirectToRoute呢？

Answer 1

redirectToRoute是你提到的Controller类的一部分。您需要做的就是自己创建方法。

首先，您需要将路由器注入CustomExceptionController(因此您需要在 DI 中将自定义控制器定义为服务(

services:
my.custom.exception_controller:
    class: CustomExceptionController
    arguments: [ "@twig", "%kernel.debug%", "@router" ]
twig:
    exception_controller: my.custom.exception_controller:showAction

您的自定义类应如下所示：

class CustomExceptionController extends SymfonyBundleTwigBundleControllerExceptionController
{
    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)
    {
    }
}

之后，您可以在 CustomExceptionController 中实现 redirectToRoute，就像在控制器中一样(或者直接创建 RedirectResponse 而无需使用帮助程序方法(

/**
 * 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

使用UrlGeneratorInterface的简单方法，例如

 getContainer()->get('router')->generate( 'custom_error', ['key' => 'some val'], 0 );

getContainer(( - 是自己的功能，请参阅手册 如果需要从服务生成 URL，请键入提示 UrlGeneratorInterface 服务：

// src/Service/SomeService.php
use SymfonyComponentRoutingGeneratorUrlGeneratorInterface;
class SomeService
{
    private $router;
    public function __construct(UrlGeneratorInterface $router)
    {
        $this->router = $router;
    }
    public function someMethod()
    {
        $url = $this->router->generate( 'custom_error', [ 'key' => 'some value' ] );
    }
}