断言错误：`HyperlinkedIdentityField`需要序列化程序内容中的请求

Question

我想建立一种many-to-many关系，一个人可以在许多俱乐部，一个俱乐部可以有许多人。我为下面的逻辑添加了models.py和serializers.py，但是当我试图在命令提示符中序列化它时，我得到以下错误-我在这里做错了什么？我甚至没有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会给我这个-

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给了我错误

**************编辑********************* 我意识到错误可能是由于url模式造成的，所以我添加了以下url模式，但仍然得到错误-

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

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 2

我有两个解决方案。。。

url.py

1） 如果使用的是router.register，则可以添加基本名称：

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

2） 如果你有这样的东西：

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

必须将上下文传递给序列化程序：

视图.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)

这样，您可以继续在序列化程序上使用url： serializers.py

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

Answer 3

由于序列化程序的context中的request预期会接收到HyperlinkedIdentityField，因此它可以生成绝对url，因此会出现此错误。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。

如果需要在命令行上检查序列化程序，则需要执行以下操作：

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

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