Question

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

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
    }
}

这不起作用，因为\ Symfony \ Bundle \ TwigBundle \ Controller \ ExceptionController不扩展类Controller。那么如何在这个类中使用$ this-＆gt; redirectToRoute？

Answer 1

redirectToRoute是您提及的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（因此您需要将自定义控制器定义为DI中的服务）

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 \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)
    {

    }
}

之后，您可以在CustomExceptionController中实现redirectToRoute，就像在Controller中完成一样（或者直接创建没有辅助方法的RedirectResponse）

/**
 * 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（如

）的简便方法

// 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' ] );

    }
}

getContainer（） - 是自己的功能，请参阅手册If you need to generate a URL from a service, type-hint the UrlGeneratorInterface service:

app/assets/fonts

访问CustomExceptionController中的redirectToRoute

2 个答案: