我在Symfony 2.3项目中使用FOSRestBundle。
我无法为响应异常设置_format。 在我的config.yml中,我有:
twig:
exception_controller: 'FOS\RestBundle\Controller\ExceptionController::showAction'
默认返回是HTML格式,但是
是否可以将_format = json
设置为返回异常?
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。
答案 0 :(得分:2)
您可以手动编写路线并在此处设置_format
:
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:或者您可以编写自己的异常处理程序,并根据需要执行以下操作:
// 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);
}
}
并将其注册为听众:
# 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 }
答案 1 :(得分:0)
在向FosRest控制器发出请求时,捕获symfony异常并返回json的最简单方法是:
# app/config/config.yml
fos_rest:
format_listener:
rules:
- { path: '^/api/', priorities: ['json', 'xml'] }
- { path: '^/', stop: true }