AssertionError: `HyperlinkedIdentityField` requires the request in the serializer context

Question

我想建立一个 many-to-many 关系，一个人可以在多个俱乐部，一个俱乐部可以有很多人。我为以下逻辑添加了 models.py 和 serializers.py 但是当我尝试在命令提示符下序列化它时，出现以下错误 - 我在这里做错了什么？我什至没有 HyperlinkedIdentityField

Traceback (most recent call last):
File "<console>", line 1, in <module>
File "C:\Users\user\corr\lib\site-packages\rest_framework\serializers.py", line 503, in data
ret = super(Serializer, self).data
File "C:\Users\user\corr\lib\site-packages\rest_framework\serializers.py", line 239, in data
self._data = self.to_representation(self.instance)
File "C:\Users\user\corr\lib\site-packages\rest_framework\serializers.py", line 472, in to_representation
ret[field.field_name] = field.to_representation(attribute)
File "C:\Users\user\corr\lib\site-packages\rest_framework\relations.py", line 320, in to_representation"the serializer." % self.__class__.__name__
AssertionError: `HyperlinkedIdentityField` requires the request in the serializer context. Add `context={'request': request}` when instantiating the serializer.

models.py

class Club(models.Model):
    club_name = models.CharField(default='',blank=False,max_length=100)

class Person(models.Model):
    person_name = models.CharField(default='',blank=False,max_length=200)
    clubs = models.ManyToManyField(Club)

serializers.py

class ClubSerializer(serializers.ModelSerializer):
    class Meta:
        model = Club
        fields = ('url','id','club_name','person')

class PersonSerializer(serializers.ModelSerializer):
    clubs = ClubSerializer()
    class Meta:
        model = Person
        fields = ('url','id','person_name','clubs')

views.py

class ClubDetail(generics.ListCreateAPIView):
serializer_class = ClubSerializer

def get_queryset(self):
     club = Clubs.objects.get(pk=self.kwargs.get('pk',None))
     persons = Person.objects.filter(club=club)
     return persons

class ClubList(generics.ListCreateAPIView):
    queryset = Club.objects.all()
    serializer_class = ClubSerializer


class PersonDetail(generics.RetrieveUpdateDestroyAPIView):
    serializer_class = PersonSerializer


def get_object(self):
    person_id = self.kwargs.get('pk',None)
    return Person.objects.get(pk=person_id)

检查创建的序列化程序给了我这个 -

PersonSerializer(<Person: fd>):
url = HyperlinkedIdentityField(view_name='person-detail')
id = IntegerField(label='ID', read_only=True)
person_name = CharField(max_length=200, required=False)
clubs = ClubSerializer():
    url = HyperlinkedIdentityField(view_name='club-detail')
    id = IntegerField(label='ID', read_only=True)
    club_name = CharField(max_length=100, required=False)

但是serializer.data给我错误

Edit

我意识到错误可能是因为 url 模式，所以我添加了以下 url 模式，但我仍然得到错误 -

urlpatterns = format_suffix_patterns([
url(r'^$', views.api_root),
url(r'^clubs/$',
    views.ClubList.as_view(),
    name='club-list'),
 url(r'^clubs/(?P<pk>[0-9]+)/persons/$',
    views.ClubDetail.as_view(),
    name='club-detail'),
url(r'^person/(?P<pk>[0-9]+)/$',
    views.PersonDetail.as_view(),
    name='person-detail'),
])

Answer 1

您收到此错误是因为 HyperlinkedIdentityField 期望在序列化程序的 context 中接收 request 以便它可以构建绝对 URL。当您在命令行上初始化序列化程序时，您无权访问请求，因此收到错误消息。

如果你需要在命令行上检查你的序列化器，你需要做这样的事情：

from rest_framework.request import Request
from rest_framework.test import APIRequestFactory

from .models import Person
from .serializers import PersonSerializer

factory = APIRequestFactory()
request = factory.get('/')


serializer_context = {
    'request': Request(request),
}

p = Person.objects.first()
s = PersonSerializer(instance=p, context=serializer_context)

print s.data

您的 url 字段看起来像 http://testserver/person/1/。

Answer 2

我遇到了同样的问题。我的方法是从 serializer.py 中的 Meta.fields 中删除 'url'。

Answer 3

我有两个解决方案...

urls.py

1) 如果您使用的是 router.register，您可以添加 base_name:

router.register(r'users', views.UserViewSet, base_name='users')
urlpatterns = [    
    url(r'', include(router.urls)),
]

2) 如果你有这样的事情：

urlpatterns = [    
    url(r'^user/$', views.UserRequestViewSet.as_view()),
]

您必须将上下文传递给序列化程序：

views.py

class UserRequestViewSet(APIView):            
    def get(self, request, pk=None, format=None):
        user = ...    
        serializer_context = {
            'request': request,
        }
        serializer = api_serializers.UserSerializer(user, context=serializer_context)    
        return Response(serializer.data)

像这样你可以继续在你的序列化器上使用url： serializers.py

...
url = serializers.HyperlinkedIdentityField(view_name="user")
...

Answer 4

根据 Slipstream 的回答，我编辑了我的 views.py 介绍上下文，现在它起作用了。

class UserViewSet(viewsets.ModelViewSet):

    """
    API endpoint that allows users to be viewed or edited.
    """
    queryset = User.objects.all().select_related('profile').order_by('-date_joined')
    serializer_class = UserSerializer

    @list_route(methods=['get'], url_path='username/(?P<username>\w+)')
    def getByUsername(self, request, username):
        serializer_context = {
            'request': request,
        }
        user = get_object_or_404(User, username=username)
        return Response(UserSerializer(user, context=serializer_context).data, status=status.HTTP_200_OK)

Answer 5

根据MDT的回复，我使用了django-rest-framework，将request改成request._request解决了。

serializer_context = {'request': Request(request._request)}

Answer 6

您可以简单地通过将实例化（在 views.py 中）更改为这样的事情来解决它：

your_serializer = YourModelSerializer(YourQuerySet_or_object, many=True,context={'request':request})

Answer 7

在您只需要相对 URL 的情况下，您可以简单地将 None 传递给 'request' 中的 context 键，例如；在命令行中测试序列化程序。

serializer = YourModelSerializer(modelInstance_or_obj, context={'request': None})

Answer 8

对于外部网址，您只需将请求放在 None:

context={
    'request': None
},

Answer 9

在我的例子中，我不得不将字段名称从 url 更改为任何其他名称。讨厌自动魔法

AssertionError: `HyperlinkedIdentityField` requires the request in the serializer context

AssertionError: `HyperlinkedIdentityField` requires the request in the serializer context

python

django

django-views

django-serializer

django-rest-framework