Question

我在我的应用程序中使用django rest框架。在视图中，对于获取查询，我以查询集的形式获取模型列表并尝试传递给我的Response（）

organizationViews.py

 def get(self,request):
    user = self.request.user
    result = OrganizationModel.object.all()
     serializer = OrganizationSerializer("json", list(result), many=True)
     return Response(serializer, status=status.HTTP_200_OK)

organizationModel.py

class OrganizationManager(models.Manager):
    def create_organization(self, _name):
        org = self.model( name = _name)
        org.save(using=self._db)
        return org


class Organization(BaseModel):
    name = models.CharField(_('organization'), max_length=100,       blank=True,unique=True)

def as_json(self):
    return dict(
        id=self.id,
        org_name=self.name)

object = OrganizationManager()

class Meta(BaseModel.Meta):
    db_table = 'organizations'
    verbose_name = _('organization')
    verbose_name_plural = _('organizations')

def __str__(self):
    return self.name

我收到以下错误：

Traceback (most recent call last):
 File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/core/handlers/base.py", line 174, in get_response
 response = self.process_exception_by_middleware(e, request)
  File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/core/handlers/base.py", line 172, in get_response
response = response.render()
  File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/django/template/response.py", line 160, in render
self.content = self.rendered_content
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/response.py", line 71, in rendered_content
ret = renderer.render(self.data, media_type, context)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/renderers.py", line 104, in render
separators=separators
 File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/json/__init__.py", line 237, in dumps
**kw).encode(obj)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/json/encoder.py", line 199, in encode
chunks = self.iterencode(o, _one_shot=True)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/json/encoder.py", line 257, in iterencode
return _iterencode(o, 0)
File "/Users/richagupta/VirtualEnvs/py35/lib/python3.5/site-packages/rest_framework/utils/encoders.py", line 64, in default
 return super(JSONEncoder, self).default(obj)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/json/encoder.py", line 180, in default
 raise TypeError(repr(o) + " is not JSON serializable")
 TypeError: OrganizationSerializer('json', [<Organization: zeotap>, <Organization: movial>, <Organization: Redbull>, <Organization: mogey>, <Organization: vizury>], many=True):
id = IntegerField()
name = CharField() is not JSON serializable

如果我在接到电话中跟进：

serialized_q= json.dumps(list(queryset), cls=DjangoJSONEncoder)
return Response(serialized_q, status=status.HTTP_200_OK)

我没有收到错误，但结果如下：

 "[{\"id\": 1, \"soft_delete\": false, \"created_at\": \"2016-04-14T13:35:21.636Z\", \"name\": \"org1\", \"updated_at\": \"2016-04-14T13:35:21.636Z\"}, 
  {\"id\": 4, \"soft_delete\": false, \"created_at\": \"2016-04-14T13:37:02.230Z\", \"name\": \"org2\", \"updated_at\": \"2016-04-14T13:37:02.230Z\"}]"

我应该如何将queryset传递给Response方法？

Answer 1

你真的不想覆盖get方法，你可能最好覆盖get_queryset;

class ViewSet(ModelViewSet):
    queryset = Model.objects.all()
    serializer_class = ModelSerializer

    def get_queryset(self): #this method is called inside of get
        queryset = self.queryset.filter(doyourfiltering=True)
        return queryset

你可以直接在这里返回一个查询集，Django会为你序列化它。您可以通过self.request访问该请求。

Answer 2

我认为您只能使用序列化程序来序列化单个模型。所以而不是：

serializer = OrganizationSerializer("json", list(result), many=True)

尝试：

data = [
    OrganizationSerializer(model).data
    for model in result
]

然后只是：

return Response(data)

请参阅：http://www.django-rest-framework.org/api-guide/serializers/#serializing-objects

Django Rest框架：将queryset传递给Response（）

2 个答案: