python - 在 Django REST 框架 API 根中包含 list_route 方法

Question

我正在使用 Django REST 框架，并且我有一个带有额外列表路由方法的 View 集。我怎样才能让该方法的 URL 包含在 API 根页面中？

这是我的 View 集的简化版本:

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)

我在 urls.py 中注册 View 集:

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

books/featured 的 URL 路由正确，但是当我转到 http://localhost:8000/api 时，我只看到这个:

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

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

我怎样才能为这样的东西添加一个条目？

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

Answer 1

可以尝试继承负责api Root View 的DefaultRouter，重新定义get_api_root_view方法。

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()

附言抱歉，在我发布答案之前没有看到您的评论