Laravel Dingo Api - 如何从API控制器返回JSON格式的错误响应？

3 回答

• 2

  在构建api时，您可以在全局范围内捕获类似的异常类型 . 例如，如果用户试图获取具有不存在的ID的客户，则可以执行此操作 .

  Customer::findOrFail($id);
  

  然后你可以像这样在 app/start/global.php 中捕获所有这种类型的异常 .

  App::error(function(ModelNotFoundException $modelNotFoundException){
      $errorResponse = [
          'errors'    => 'Not found any resource',
          'message'   => $modelNotFoundException->getMessage()
      ];
  
      return Response::json($errorResponse, 404);     //404 = Not found
  });
  

  回复于 2024-04-28T03:14:06+08:00
• 0

  阅读Dingo的Returning Errors文档，它说：

  而不是手动创建和返回错误响应，您可以简单地抛出异常，包将处理异常并返回适当的响应 . 以下是遇到错误时应抛出的所有受支持异常的列表 . Symfony \ Component \ HttpKernel \ Exception \ AccessDeniedHttpException Symfony \ Component \ HttpKernel \ Exception \ BadRequestHttpException Symfony \ Component \ HttpKernel \ Exception \ ConflictHttpException Symfony \ Component \ HttpKernel \ Exception \ GoneHttpException Symfony \ Component \ HttpKernel \ Exception \ HttpException Symfony \ Component \ HttpKernel \ Exception \ LengthRequiredHttpException Symfony \ Component \ HttpKernel \ Exception \ MethodNotAllowedHttpException Symfony \ Component \ HttpKernel \ Exception \ NotAcceptableHttpException Symfony \ Component \ HttpKernel \ Exception \ NotFoundHttpException Symfony \ Component \ HttpKernel \ Exception \ PreconditionFailedHttpException Symfony \ Component \ HttpKernel \ Exception \ PreconditionRequiredHttpException Symfony \ Component \ HttpKernel \ Exception \ ServiceUnavailableHttpException Symfony \ Component \ HttpKernel \ Exception \ TooManyRequestsHttpException Symfony \ Component \ HttpKernel \ Exception \ UnauthorizedHttpException Symfony \ Component \ HttpKernel \ Exception \ UnsupportedMediaTypeHttpException

  它还支持一些通用资源异常，您也可以将验证错误传递到：

  Dingo \ Api \ Exception \ DeleteResourceFailedException Dingo \ Api \ Exception \ ResourceException Dingo \ Api \ Exception \ StoreResourceFailedException Dingo \ Api \ Exception \ UpdateResourceFailedException

  所以简而言之，你需要抛出一个上面的例外，Dingo支持回到Dingo . 例如：

  try {
      //Some code that returns an exception
  } catch(SomeException $e) {
      throw new Symfony\Component\HttpKernel\Exception\HttpException($e->getMessage);
  }
  

  或者实际上，如果抛出的异常是上面的类型之一，或者扩展它们的异常，你可以完全删除你的try / catch子句 . 异常应该自动抛回Dingo来处理它 .

  回复于 2024-04-28T03:14:06+08:00
• 2

  请检查一下：

  try {
      //some code that returns an exception
  } catch(\Exception $e) {
      $response = array(
          'message' => 'some random exception message'
      );
      return response()->json($response, 403);
  }
  

  请检查并让我知道 .

  回复于 2024-04-28T03:14:06+08:00