custom exception behavior in symfony2

Question

Im trying to figure out how to make a custom exception behavior. When i throw a exception using

 throw new \Exception('Error occurred with your request please try again');

I automatically get status 500 and the message as internal server error

However i would instead like my response to include my exception message instead of just internal server error so for it to display something like so:

 {
   "error":{
      "code":500, 
      "message":"Error occurred with your request please try again"
   }
 }

and on top of that possibly do some extra things such as email myself the error. However I only want this to happen when i throw a \Exception as opposed to using something like

    throw new HttpException

Any help or ideas on how to accomplish this.

I should also mention that I am not using Twig or any templates for this. This is strictly a API type response

Answer 1

Take a look at http://symfony.com/doc/current/cookbook/controller/error_pages.html There is enough information to get you started.

In short, you should create app/Resources/TwigBundle/views/Exception/exception.json.twig and there you have access to the exception.message and error_code.

here's solution for you:

{% spaceless %}
{
  "error":{
    "code": {{ error_code }}, 
    "message":{{ exception.message }}
  }
}
{% endspaceless %}

Another solution is to use Exception Listener:

namespace Your\Namespace;

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

class JsonExceptionListener
{
    public function onKernelException(GetResponseForExceptionEvent $event)
    {
        $exception = $event->getException();
        $data = array(
            'error' => array(
                'code' => $exception->getCode(),
                'message' => $exception->getMessage()
            )
        );
        $response = new JsonResponse($data, $exception->getCode());
        $event->setResponse($response);
    }
}

update your services config:

json_exception_listener:
    class: Your\Namespace\JsonExceptionListener
    tags:
        - { name: kernel.event_listener, event: kernel.exception, method: onKernelException, priority: 200 }

cheers

Answer 2

If you want to be able to read the message you are sending back, you need to return a response code that returns text with the response, like 200. So you can do something like this with a try catch block:

try{
    //.....
    throw new \Exception('Error occurred with your request please try again');
}catch(\Exception $ex){
    $return = array('error'=>array('code'=>500,'message'=>$ex->getMessage()));
    return new Response(json_encode($return),200,array('Content-Type' => 'application/json'));
}

and on the client side, you should get a json object exactly as you outlined you want it.

even better, you can do different things depending on which exception is thrown

}catch(\Exception $ex){
    $class = get_class($ex);
    if($class == 'Symfony\Component\HttpKernel\Exception\HttpException'){
        // create a response for HttpException
    }else{
        // create a response for all other Exceptions
    }
}