Django Rest Framework：序列化程序响应错误

Question

我在 DRF 中的视图集有问题：

def get_queryset(self):
    """
    :return: filtered queryset based on request query_param

    """
    # todo#amedeo: improve the readability code
    qs = super(ChecklistViewSet, self).get_queryset()
    _pk = self.kwargs['pk']

    if self.request.method == 'PUT':
        instance = qs.filter(pk=int(_pk)).first()
        # pass in the instance we want to update
        serializer = ChecklistSerializer(instance, self.request.data)

        # validate and update
        if serializer.is_valid():
            serializer.save()
            serializer_dict = serializer.data
            serializer_dict["message"] = "Checklist updated successfully."
            return response.Response(serializer_dict, status=status.HTTP_200_OK)

        else:
            return response.Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

在我的代码中保存了 objetc，但响应给出了一个错误：

AttributeError：“响应”对象没有属性“模型”

我的序列化器是：

class ChecklistSerializer(serializers.ModelSerializer):

class Meta:
    model = Checklist
    fields = ('id', 'approved_by', 'role', 'hardship_classification',
              'total_capacity', 'capacity_for_residents', 'checklist_type',
              'state', 'pdf', 'submitting_afp', 'disabled', 'location')

我通过传递 json 的客户端 PUT 方法调用：

{
"approved_by": "Test",
"role": "test_role_4",
"hardship_classification": "test_6",
"total_capacity": "50",
"capacity_for_residents": "350",
"checklist_type": "permanent",
"state": "qcl_draft",
"pdf": null,
"submitting_afp": 3999,
"disabled": false,
"location": 97
}

但我收到了那个错误。模型已保存，但响应给了我错误

内部服务器错误：/api/v1/checklists/3/ 回溯（最近一次通话最后）： 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/django/core/handlers/exception.py”，第 41 行，在内部 响应 = get_response（请求） _get_response 中的文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/django/core/handlers/base.py”，第 187 行 response = self.process_exception_by_middleware(e, request) _get_response 中的文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/django/core/handlers/base.py”，第 185 行 响应 = Wrapped_callback(request, *callback_args, **callback_kwargs) 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/django/views/decorators/csrf.py”，第 58 行，位于 Wrapped_view 返回 view_func(*args, **kwargs) 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/viewsets.py”，第 116 行，在视图中 return self.dispatch(request, *args, **kwargs) 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/views.py”，第 495 行，在调度中 响应 = self.handle_exception(exc) 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/views.py”，第 455 行，在 handle_exception self.raise_uncaught_exception(exc) 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/views.py”，第 483 行，在调度中 self.initial（请求，*args，**kwargs） 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/views.py”，第 401 行，初始 self.check_permissions(请求) 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/views.py”，第 334 行，在 check_permissions 如果不是 permission.has_permission(request, self): 文件“/home/adifilippo/.envs/accomodation/local/lib/python2.7/site-packages/rest_framework/permissions.py”，第 206 行，在 has_permission perms = self.get_required_permissions(request.method, queryset.model) AttributeError：“响应”对象没有属性“模型” "PUT /api/v1/checklists/3/HTTP/1.1" 500 42934

在此处输入代码

Answer 1

get_queryset 方法应该返回 queryset 而不是 Response，并且更新操作应该在 put 方法中被覆盖。阅读此处class-based-views 和 filtering 了解更多详情。可能您需要删除 get_queryset 覆盖并添加 put 方法，如下所示：

def put(self, request, pk, format=None):
    instance = self.get_object(pk)
    # pass in the instance we want to update
    serializer = ChecklistSerializer(instance, self.request.data)

    # validate and update
    if serializer.is_valid():
        serializer.save()
        serializer_dict = serializer.data
        serializer_dict["message"] = "Checklist updated successfully."
        return response.Response(serializer_dict, status=status.HTTP_200_OK)

    else:
        return response.Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

Answer 2

你可以这样做

from django.shortcuts import get_objects_or_404

class YourAPIView(ModelViewSet):
   def put(self, request, pk):
       instance = get_objects_or_404(YourModel, pk=pk)
       serializer = ChecklistSerializer(instance, self.request.data)

# validate and update
       if serializer.is_valid():
           serializer.save()
           serializer_dict = serializer.data
           serializer_dict["message"] = "Checklist updated successfully."
           return response.Response(serializer_dict, status=status.HTTP_200_OK)
       return response.Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)