异常Eventlistener没有捕获RabbitMQ使用者的异常

Question

I'm creating a Microservice in Symfony where I need to use RabbitMQ to consume messages to create a document. The consumer is working but whenever a Exception is thrown I want to log the Exception in a format and in a separate log file.

I'm using Symfony 4.2 and all Exceptions that are thrown are logged in the dev.log. I made a new channel in Monolog and called in exception, so that I can manage my loggings.

For this manner I am using an ExceptionListener and I followed the instructions from the Symfony Docs.

I have my Consumer setup as follows

public function execute(AMQPMessage $msg): int
    {
        try {
            return $this->documentService->createDocument($this->serializer->deserialize($msg->body, 'array', 'json'));
        } catch (\Exception $e) {
             $this->logger->error($e);

            return ConsumerInterface::MSG_REJECT;
        }
    }

And inside the service I have the following:

try {
...

    $template = $this->templateRepository->findOneByName($messageData['template']);
    if ($template === null) throw new \Exception('Template does not exist');

...

    return ConsumerInterface::MSG_ACK;
} catch (\Exception $e) {
    $this->logger->error($e);

    return ConsumerInterface::MSG_REJECT;
}

Im logging on both classes but when an Exceptions throws I get the Exception in the console but the Listener does not catches the ExceptionEvent.

I have tried different events and I added http-kernel package to the Symfony, since I am using Symfony Flex I don't have all packages.

I have registered my Listener in services.yaml as following:

    App\EventListener\DocumentExceptionListener:
        tags:
            - { name: kernel.event_listener, event: kernel.exception, method: onKernelException }
        arguments:
            $logger: '@monolog.logger.exception' # Log all exceptions to a separate log file

And my ExceptionListener is as following:

public function __construct(LoggerInterface $logger, SerializerInterface $serializer)
    {
        $this->logger = $logger;
        $this->serializer = $serializer;

        $this->logger->info('Starting to listen...');
    }

    public function onKernelException(GetResponseForExceptionEvent $event)
    {
        $this->logger->info('currently listening...');

        $this->log($event->getException());
    }

    private function log(\Exception $exception)
    {
        $log = [
            'code' => $exception->getCode(),
            'message' => $exception->getMessage(),
            'called' => [
                'file' => $exception->getTrace()[0]['file'],
                'line' => $exception->getTrace()[0]['line'],
            ],
            'occurred' => [
                'file' => $exception->getFile(),
                'line' => $exception->getLine(),
            ],
        ];

        if ($exception->getPrevious() instanceof Exception) {
            $log += [
                'previous' => [
                    'message' => $exception->getPrevious()->getMessage(),
                    'exception' => get_class($exception->getPrevious()),
                    'file' => $exception->getPrevious()->getFile(),
                    'line' => $exception->getPrevious()->getLine(),
                ],
            ];
        }

        $this->logger->error($this->serializer->serialize($log, 'json'));
    }

My output right now is:

exception.INFO: Starting to listen... [] []

Answer 1

I read the word "console" in your question. If it is a console exception (because it is about a command execution) you have to tag your listener another way:

tags:
    - { name: kernel.event_listener, event: console.error }

See the console.error event name here.

See more here: Events and Event Listeners