在 obj_create 方法中返回美味派的自定义消息答案

【问题标题】：Returning custom message in tastypie in obj_create method在 obj_create 方法中返回美味派的自定义消息
【发布时间】：2015-02-04 12:59:54
【问题描述】：

我正在tastepie中编写Rest API并创建如下自定义资源

class Myresource(Resource):
      def obj_create(self, request, **kwargs):
          # logic when POST request is called
          # here if some error occurs while inserting data I want to add my custom error message and return it response. 
          return bundle

如何从 obj_create 消息返回自定义错误消息。现在我正在处理异常，但 sweetpie 总是返回 201。

【问题讨论】：

标签： python tastypie

【解决方案1】：

我认为您需要重写资源中的 wrap_view 方法，如 here 所述。

这就是它的工作方式。

class YourResource(ModelResource):

def wrap_view(self, view):
    """
    Wraps views to return custom error codes instead of generic 500's
    """
    @csrf_exempt
    def wrapper(request, *args, **kwargs):
        try:
            callback = getattr(self, view)
            response = callback(request, *args, **kwargs)

            if request.is_ajax():
                patch_cache_control(response, no_cache=True)

            # response is a HttpResponse object, so follow Django's instructions
            # to change it to your needs before you return it.
            # https://docs.djangoproject.com/en/dev/ref/request-response/
            return response
        except (BadRequest, ApiFieldError), e:
            return HttpBadRequest({'code': 666, 'message':e.args[0]})
        except ValidationError, e:
            # Or do some JSON wrapping around the standard 500
            return HttpBadRequest({'code': 777, 'message':', '.join(e.messages)})
        except Exception, e:
            # Rather than re-raising, we're going to things similar to
            # what Django does. The difference is returning a serialized
            # error message.
            return self._handle_500(request, e)

    return wrapper

【讨论】：