Symfony 3 API REST - 尝试捕获 JSON 响应格式的异常

Question

我使用 Symfony 和这些捆绑包 FosRestBundle、jms/serializer-bundle、lexik/jwt-authentication-bundle 创建 api rest 服务器。

我怎样才能像这样发送一个干净的 json 响应格式：

Missing field "NotNullConstraintViolationException"
    {'status':'error','message':"Column 'name' cannot be null"}
or
    {'status':'error','message':"Column 'email' cannot be null"}
Or Duplicate entry "UniqueConstraintViolationException" :
    {'status':'error','message':"The email user1@gmail.com exists in database."}

代替系统消息：

UniqueConstraintViolationException in AbstractMySQLDriver.php line 66: An exception occurred while executing 'INSERT INTO user (email, name, role, password, is_active) VALUES (?, ?, ?, ?, ?)' with params ["user1@gmail.com", "etienne", "ROLE_USER", "$2y$13$tYW8AKQeDYYWvhmsQyfeme5VJqPsll\/7kck6EfI5v.wYmkaq1xynS", 1]: SQLSTATE[23000]: Integrity constraint violation: 1062 Duplicate entry 'user1@gmail.com' for key 'UNIQ_8D93D649E7927C74'

返回一个干净的 json 响应，名称为必填字段或遗漏字段。

这里是我的控制器：

    <?php

namespace AppBundle\Controller;

use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Method;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use AppBundle\Entity\User;
use Doctrine\DBAL\Exception\UniqueConstraintViolationException;
use Symfony\Component\Debug\ExceptionHandler;
use Symfony\Component\Debug\ErrorHandler;
use Symfony\Component\HttpFoundation\JsonResponse;

use FOS\RestBundle\Controller\Annotations as Rest; // alias pour toutes les annotations

class DefaultController extends Controller
{


 /**
 * @Rest\View()
 * @Rest\Post("/register")
 */
public function registerAction(Request $request)
{
    //catch all errors and convert them to exceptions
    ErrorHandler::register();

    $em = $this->get('doctrine')->getManager();
    $encoder = $this->container->get('security.password_encoder');

    $username = $request->request->get('email'); 
    $password = $request->request->get('password');
    $name = $request->request->get('name');

    $user = new User($username);

    $user->setPassword($encoder->encodePassword($user, $password));
    $user->setName($name);
    $user->setRole('ROLE_USER');
    try {
        $em->persist($user);
        $em->flush($user);
    }
    catch (NotNullConstraintViolationException $e) {
        // Found the name of missed field
            return new JsonResponse();
    } 
    catch (UniqueConstraintViolationException $e) {
        // Found the name of duplicate field
            return new JsonResponse();
    } 
    catch ( \Exception $e ) {

        //for debugging you can do like this
        $handler = new ExceptionHandler();
        $handler->handle( $e );
        return new JsonResponse(
            array(
                'status' => 'errorException', 
                'message' => $e->getMessage()
            )
        );
    }

    return new Response(sprintf('User %s successfully created', $user->getUsername()));
}
}

谢谢

Answer 1

我们使用以下方法：

API 异常的通用类：

class ApiException extends \Exception
{  
    public function getErrorDetails()
    {
        return [
            'code' => $this->getCode() ?: 999,
            'message' => $this->getMessage()?:'API Exception',
        ];
    }
}

创建一个扩展 ApiException 的验证异常

class ValidationException extends ApiException
{
    private $form;

    public function __construct(FormInterface $form)
    {
        $this->form = $form;
    }

    public function getErrorDetails()
    {
        return [
            'code' => 1,
            'message' => 'Validation Error',
            'validation_errors' => $this->getFormErrors($this->form),
        ];
    }

    private function getFormErrors(FormInterface $form)
    {
        $errors = [];
        foreach ($form->getErrors() as $error) {
            $errors[] = $error->getMessage();
        }
        foreach ($form->all() as $childForm) {
            if ($childForm instanceof FormInterface) {
                if ($childErrors = $this->getFormErrors($childForm)) {
                    $errors[$childForm->getName()] = $childErrors;
                }
            }
        }

        return $errors;
    }
}

当表单出现错误时在控制器中使用异常

if ($form->getErrors(true)->count()) {
    throw new ValidationException($form);
}

创建和配置您的 ExceptionController

class ExceptionController extends FOSRestController
{

    public function showAction($exception)
    {
        $originException = $exception;

        if (!$exception instanceof ApiException && !$exception instanceof HttpException) {
            $exception = new HttpException($this->getStatusCode($exception), $this->getStatusText($exception));
        }

        if ($exception instanceof HttpException) {
            $exception = new ApiException($this->getStatusText($exception), $this->getStatusCode($exception));
        }

        $error = $exception->getErrorDetails();

        if ($this->isDebugMode()) {
            $error['exception'] = FlattenException::create($originException);
        }

        $code = $this->getStatusCode($originException);

        return $this->view(['error' => $error], $code, ['X-Status-Code' => $code]);
    }

    protected function getStatusCode(\Exception $exception)
    {
        // If matched
        if ($statusCode = $this->get('fos_rest.exception.codes_map')->resolveException($exception)) {
            return $statusCode;
        }

        // Otherwise, default
        if ($exception instanceof HttpExceptionInterface) {
            return $exception->getStatusCode();
        }

        return 500;
    }

    protected function getStatusText(\Exception $exception, $default = 'Internal Server Error')
    {
        $code = $this->getStatusCode($exception);

        return array_key_exists($code, Response::$statusTexts) ? Response::$statusTexts[$code] : $default;
    }

    public function isDebugMode()
    {
        return $this->getParameter('kernel.debug');
    }
}

config.yml

fos_rest:
    #...
    exception:
        enabled: true
        exception_controller: 'SomeBundle\Controller\ExceptionController::showAction'

见：http://symfony.com/doc/current/bundles/FOSRestBundle/4-exception-controller-support.html

使用这种方法可以为每种类型的错误创建带有自定义消息和代码的自定义异常（有助于 API 文档），另一方面隐藏其他内部异常，仅在抛出异常时向 API 使用者显示“内部服务器错误”不是从 APIException 扩展而来的。