Django Rest required parameters in URL

Question

I'm using django rest framework. Here is my code:

urls.py:

urlpatterns = [
    url(r'^users/show', UserShow.as_view()),
]

view.py:

class UserShow(ListAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

    def get_queryset(self):
        queryset = User.objects.all()
        username = self.request.query_params.get('username', None)
        user_id = self.request.query_params.get('user_id', None)
        if username is not None:
            queryset = queryset.filter(username=username)
        if user_id is not None:
            queryset = queryset.filter(pk=user_id)
        return queryset

I want to get values from url like this: /users/show?user_id=1 or /users/show?username=mike.

Either an user_id or username must be required parameter. How can I control it in class based views?

With my code if I'm sending the request with wrong parameter name /users/show?user111name=mike or simple /users/show the view of course response me with queryset = User.objects.all() and lists all the users. I don't need that. I need if required parameters are None response with 404.

I can get needed result with function based view:

@api_view(['GET'])
def users(request):
    if request.method == 'GET':
        queryset = User.objects.all()
        username = request.GET.get('username', None)
        user_id = request.GET.get('user_id', None)

        if username is not None:
            queryset = queryset.filter(username=username)
        elif user_id is not None:
            queryset = queryset.filter(pk=user_id)
        else:
            return Response({"status": "required field not found."},
                            status=status.HTTP_404_NOT_FOUND)

        if not queryset.exists():
            return Response({"status": "not found."},
                            status=status.HTTP_404_NOT_FOUND)

        serializer = UserSerializer(queryset, many=True)
        return Response(serializer.data)

But how can I do it with generic class based views?

You should take a look at stackoverflow.com/questions/36505792/… — trinchet
– trinchet, Commented May 27, 2016 at 13:49
Could you please explain to me why you are using ListAPIView to get single object? There is a RetrieveAPIView dedicated for such a use, plus you can have username or id used as part of the url not query.. — Jerzyk
– Jerzyk, Commented Jun 5, 2016 at 1:37

Jerzyk · Accepted Answer · 2023-03-16 06:47:58Z

4

class UserIdRetrieve(RetrieveAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer
    
class UserUsernameRetrieve(UserIdRetrieve):
     lookup_field = 'username'

and in urls:

urlpatterns = [
    url(r'^users/(?P<pk>\d+)/', UserIdRetrieve.as_view()), 
    url(r'^users/by-username/(?P<username>\w+)/', UserUsernameRetrieve.as_view())
]

if your url structure is a must, small change to above:

class UserIdRetrieve(RetrieveAPIView):
    queryset = User.objects.all()
    serializer_class = UserSerializer

    def get_object(self):
        queryset = self.filter_queryset(self.get_queryset())

        if 'username' in self.request.query_params:
            filter_kwargs = {'username': self.request.query_params['username']}
        elif 'user_id' in self.request.query_params:
             filter_kwargs = {'id': self.request.query_params['user_id']}
        else:
            raise ValidationError('Missing required parameters')

        obj = get_object_or_404(queryset, **filter_kwargs)

        # May raise a permission denied
        self.check_object_permissions(self.request, obj)

        return obj

and in urls:

urlpatterns = [
    url(r'^users/show', UserRetrieve.as_view())
]

edited Mar 16, 2023 at 6:47

answered Jun 5, 2016 at 1:47

Jerzyk

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3 Comments

arstj Over a year ago

I'm using ListAPIView because RetrieveAPIView requires pk as argument but in my case I'm not getting it from url in classic django way, due to using query_params. Of course your variant will work, but take a look at required url format again, it's not exactly what I need.

Jerzyk Over a year ago

added version that matches your url structure

Montaro Over a year ago

I would return 400 Bad Request response in case of missing required parameter not 404

Anoop · Accepted Answer · 2016-05-27 15:37:54Z

2

class UserShow(ListAPIView):

    queryset = User.objects.all()
    serializer_class = UserSerializer

    def filter_queryset(self, queryset):
        username = self.request.query_params.get('username', None)
        user_id = self.request.query_params.get('user_id', None)

        if username is not None:
            queryset = queryset.filter(username=username)
        if user_id is not None:
            queryset = queryset.filter(pk=user_id)
        return queryset

    def list(self,request,*args,**kwargs):
        username = self.request.query_params.get('username', None)
        user_id = self.request.query_params.get('user_id', None)
        if not (username or user_id):
            return Response({"status": "Required field not found."},
                                        status=status.HTTP_404_NOT_FOUND)
        return super(UserShow, self).list(request,*args,**kwargs)

answered May 27, 2016 at 15:37

Anoop

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2 Comments

arstj Over a year ago

It's working but is there more DRY way? Anyway I'll mark your answer as solution for now.

Jerzyk Over a year ago

ListAPIView should not be used for retrieving single instance

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