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

Question

I want to create a many-to-many relationship where one person can be in many clubs and one club can have many persons. I added the models.py and serializers.py for the following logic but when I try to serialize it in the command prompt, I get the following error – What am I doing wrong here? I don’t even have a HyperlinkedIdentityField

Traceback (most recent call last):
File "<console>", line 1, in <module>
File "C:Usersusercorrlibsite-packagesrest_frameworkserializers.py", line 503, in data
ret = super(Serializer, self).data
File "C:Usersusercorrlibsite-packagesrest_frameworkserializers.py", line 239, in data
self._data = self.to_representation(self.instance)
File "C:Usersusercorrlibsite-packagesrest_frameworkserializers.py", line 472, in to_representation
ret[field.field_name] = field.to_representation(attribute)
File "C:Usersusercorrlibsite-packagesrest_frameworkrelations.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)

Inspecting the created serializer gives me this –

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)

but serializer.data gives me the error

Edit

I realized the error could be because of url patterns, so I added the following url patterns but I still get the error –

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'),
])

Asked By: qwertp

||

Source

Answer 1

You’re getting this error as the HyperlinkedIdentityField expects to receive request in context of the serializer so it can build absolute URLs. As you are initializing your serializer on the command line, you don’t have access to request and so receive an error.

If you need to check your serializer on the command line, you’d need to do something like this:

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

Your url field would look something like http://testserver/person/1/.

Answered By: Ashley 'CptLemming' Wilson

Answer 2

I came across the same problem. My approach is to remove ‘url’ from Meta.fields in serializer.py.

Answered By: Diansheng

Answer 3

I have two solutions…

urls.py

1)
If you are using a router.register, you can add the base_name:

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

2)
If you have something like this:

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

You have to pass the context to the serializer:

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)

Like this you can continue to use the url on your serializer:
serializers.py

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

Answered By: Slipstream

Answer 4

Following Slipstream’s answer, I edited my views.py introducing the context and now it works.

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)

Answered By: MDT

Answer 5

Following MDT’s response, I use django-rest-framework, and solve it by changing request to request._request.

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

Answered By: Chestermomo

Answer 6

You may simply solve it by changing the instantiation (in views.py) to thing like this:

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

Answered By: Mahrez BenHamad

Answer 7

You can simply pass None to 'request' key in context in situations where you just need the relative URL, e.g; testing a serializer in command line.

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

Answered By: Pooya Kamranjam

Answer 8

For externals urls you can simply put request at None:

context={
    'request': None
},

Answered By: Yoann GUILLARD

Answer 9

In my case I had to change a field’s name from url to any other thing. Hate automagic

Answered By: juan Isaza

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

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