[Django]-Django REST framework foreign keys and filtering

by

12👍

✅

urls.py

url('^model/by/(?P<make>\w+)/$', ModelByMakerList.as_view()),

views.py

class ModelByMakerList(generics.ListAPIView):
    serializer_class = ModelSerializer

    def get_queryset(self):
        """
        This view should return a list of all models by
        the maker passed in the URL
        """
        maker = self.kwargs['make']
        return Model.objects.filter(make=maker)

For more info checkout the docs.

You can also use filtering with QUERY_PARAMS, but IMHO this looks better.

👤danielcorreia

44👍

You can specify filter_fields = ('make__slug', ) in your view set. Don’t forget to include filter_backends = (DjangoFilterBackend, ) as well. Also you will need to add django-filter dependency.

class ModelViewSet(viewsets.ModelViewSet):
    queryset = Model.objects.all()
    serializer_class = ModelSerializer
    filter_backends = (filters.DjangoFilterBackend,)
    filter_fields = ('make__slug',)

Then you query like /api/models/?make__slug=ford. Note double underscore symbol.

Docs.

If you don’t like make__slug keyword argument in the URL, then you can create a filter class:

import django_filters

from myapp.models import Make


class ModelFilter(django_filters.FilterSet):
    make = django_filters.ModelChoiceFilter(field_name="make__slug",
                                            queryset=Make.objects.all())

    class Meta:
        model = Model
        fields = ('make',)

and then

class ModelViewSet(viewsets.ModelViewSet):
    make = MakeSerializer
    queryset = Model.objects.all()
    serializer_class = ModelSerializer
    filter_backends = (filters.DjangoFilterBackend,)
    filter_class = ModelFilter

/api/models/?make=ford should work.

👤Vladimir Prudnikov

3👍

To expand on @vladimir-prudnikov’s answer:

Things changed a bit in recent versions of django-filter. You probably want:

class ModelFilter(django_filters.FilterSet):
    make = django_filters.ModelChoiceFilter(field_name='make__slug',
                                            to_field_name='slug',
                                            queryset=Make.objects.all())

    class Meta:
        model = Model
        fields = ('make',)

See https://django-filter.readthedocs.io/en/master/ref/filters.html#field-name and https://django-filter.readthedocs.io/en/master/ref/filters.html#to-field-name

👤felixhummel

1👍

What you need to do in your view is something like this:
It is called “Lookups that span relationships”

queryset = Model.objects.filter(make__name__exact='Alfa Romeo')

the filtering of models with specific engine capacity is similar

queryset = Model.objects.filter(engine_capacity__exact=5)

if you want both filters combined, you can chain them:

queryset = Model.objects.filter(make__name__exact='Alfa Romeo').filter(engine_capacity__exact=5)

more examples can be found here django query making

👤AHaberl

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