设置FOSRestBundle异常_format

Question

我在我的Symfony 2.3项目中使用FOSRestBundle。

我无法为响应异常设置_format。 在我config.yml我：

twig: 
    exception_controller: 'FOS\RestBundle\Controller\ExceptionController::showAction' 

默认返回的是HTML格式，但 是有可能设置_format = json返回例外？

我有多个捆绑包，但只有一个是RestBundle，所以其他捆绑包需要以正常方式设置。

Answer 1

您可以手动编写你的路由，并设置_format有这样的：

acme_demo.api.user: 
    type: rest 
    pattern: /user/{username_canonical}.{_format} 
    defaults: { _controller: 'AcmeDemoBundle:User:getUser', username_canonical: null, _format: json } 
    requirements: 
     _method: GET 

编辑：或者你可以写自己的异常处理程序，并与任何你需要做的例外做到：

// src/Acme/DemoBundle/EventListener/AcmeExceptionListener.php 
namespace Acme\DemoBundle\EventListener; 

use Symfony\Component\HttpKernel\Event\GetResponseForExceptionEvent; 
use Symfony\Component\HttpFoundation\JsonResponse; 
use Symfony\Component\HttpKernel\Exception\HttpExceptionInterface; 

class AcmeExceptionListener 
{ 
    public function onKernelException(GetResponseForExceptionEvent $event) 
    { 
     // do whatever tests you need - in this example I filter by path prefix 
     $path = $event->getRequest()->getRequestUri(); 
     if (strpos($path, '/api/') === 0) { 
      return; 
     } 

     $exception = $event->getException(); 
     $response = new JsonResponse($exception, 500); 

     // HttpExceptionInterface is a special type of exception that 
     // holds status code and header details 
     if ($exception instanceof HttpExceptionInterface) { 
      $response->setStatusCode($exception->getStatusCode()); 
      $response->headers->replace($exception->getHeaders()); 
     } 

     // Send the modified response object to the event 
     $event->setResponse($response); 
    } 
} 

并将其注册为一个监听器：

# app/config/config.yml 
services: 
    kernel.listener.your_listener_name: 
     class: Acme\DemoBundle\EventListener\AcmeExceptionListener 
     tags: 
      - { name: kernel.event_listener, event: kernel.exception, method: onKernelException } 

How to create an Event Listener