CODEWITHSUNDEEP
pythondjangoresturl-routing
Don Kirkby
Don Kirkby

Reputation: 56650

Include list_route methods in Django REST framework's API root

I'm using the Django REST framework, and I have a view set with an extra list route method. How can I get that method's URL included in the API root page?

Here's a simplified version of my view set:

class BookViewSet(viewsets.ReadOnlyModelViewSet):
    queryset = Book.objects.all()
    serializer_class = BookSerializer
    permission_classes = (permissions.IsAuthenticated, )

    @list_route(methods=['get'])
    def featured(self, request):
        queryset = self.filter_queryset(self.get_queryset()).filter(featured=True)

        page = self.paginate_queryset(queryset)
        if page is not None:
            serializer = self.get_serializer(page, many=True)
            return self.get_paginated_response(serializer.data)

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

I register the view set in urls.py:

router = DefaultRouter()
router.register('books', BookViewSet)
urlpatterns = patterns(
    url(r'^api/', include(router.urls), name='api_home'),
    #...
    )

The URL for books/featured is routed properly, but when I go to http://localhost:8000/api, I only see this:

HTTP 200 OK
Content-Type: application/json
Vary: Accept
Allow: GET, HEAD, OPTIONS

{
    "books": "http://localhost:8000/api/books/"
}

How can I get an entry added for something like this?

"book-featured-list": "http://localhost:8000/api/books/featured"

Upvotes: 11

Views: 6236

Answers (4)

Viktar K
Viktar K

Reputation: 1528

I use this solution based on likeon's answer:

class MyRouter(routers.DefaultRouter):
    def get_api_root_view(self, api_urls=None):
        api_root_dict = OrderedDict()
        list_name = self.routes[0].name
        for prefix, viewset, basename in self.registry:
            api_root_dict[prefix] = list_name.format(basename=basename)

        api_root_dict['books/featured'] = 'book-featured'

        return self.APIRootView.as_view(api_root_dict=api_root_dict)

Upvotes: 0

gzerone
gzerone

Reputation: 2227

You can install package django-rest-swagger, just follow here: https://github.com/marcgibbons/django-rest-swagger

It is more powerful than DRF's api list page. It will list all the rest apis (include list_route/detail_route apis) for your modules, and you can also do some api test (CRUD) on the page.

Upvotes: 2

likeon
likeon

Reputation: 791

You could try to inherit from DefaultRouter which is responsible for api root view and redefine get_api_root_view method.

class MyRouter(routers.DefaultRouter):
    def get_api_root_view(self):
        """
        Return a view to use as the API root.
        """
        api_root_dict = OrderedDict()
        list_name = self.routes[0].name
        for prefix, viewset, basename in self.registry:
            api_root_dict[prefix] = list_name.format(basename=basename)

        class APIRoot(views.APIView):
            _ignore_model_permissions = True

            def get(self, request, *args, **kwargs):
                ret = OrderedDict()
                namespace = request.resolver_match.namespace
                for key, url_name in api_root_dict.items():
                    if namespace:
                        url_name = namespace + ':' + url_name
                    try:
                        ret[key] = reverse(
                            url_name,
                            args=args,
                            kwargs=kwargs,
                            request=request,
                            format=kwargs.get('format', None)
                        )
                    except NoReverseMatch:
                        # Don't bail out if eg. no list routes exist, only detail routes.
                        continue

                ret['book-featured-list'] = '%s%s' % (ret['books'], 'featured/')

                return Response(ret)

        return APIRoot.as_view()

P.S. sorry, didn't see your comment before I posted the answer

Upvotes: 2

Marin
Marin

Reputation: 1121

Can you try with, put this below your urlpatterns. Your url will be like this: 

http://localhost:8000/api/books/

from rest_framework.urlpatterns import format_suffix_patterns

urlpatterns = format_suffix_patterns(urlpatterns)

Upvotes: -1

Related Questions