Django REST Framework中的嵌套资源

时间:2014-10-20 04:53:53

标签: python django rest django-rest-framework nested-resources

我希望使用嵌套资源实现我的新API。

Example: /api/users/:user_id/posts/

将评估特定用户的所有帖子。我还没有看到这个用例的工作示例,也许这不是实现rest API的正确方法?

3 个答案:

答案 0 :(得分:21)

正如Danilo所评论的那样,@link装饰器已移除,取而代之的是@list_route@detail_route装饰器。

更新:@detail_route& @list_route已弃用@action

以下是替代解决方案:

解决方案1:

@detail_route()
def posts(self, request, pk=None):
    owner = self.get_object()
    posts = Post.objects.filter(owner=owner)

    context = {
        'request': request
    }

    post_serializer = PostSerializer(posts, many=True, context=context)
    return Response(post_serializer.data)

解决方案2:

试试drf-nested-routers。尚未尝试过,但看起来很有希望,许多人已经在使用它。看起来像我们已经尝试实现的高级版本。

希望这有帮助。

答案 1 :(得分:4)

要映射/api/users/:user_id/posts/,您可以使用posts

ViewSet内修饰@link()方法
from rest_framework.decorators import link


class UserViewSet(viewsets.ModelViewSet):
    model = User
    serializer_class = UserSerializer

    # Your regular ModelViewSet things here

    # Add a decorated method like this
    @link()
    def posts(self, request, pk):
        # pk is the user_id in your example

        posts = Post.objects.filter(owner=pk)    

        # Or, you can also do a related objects query, something like:
        # user = self.get_object(pk)
        # posts = user.post_set.all()

        # Then just serialize and return it!
        serializer = PostSerializer(posts)
        return Response(serializer.data)

答案 2 :(得分:2)

早于commented by Danilo Cabello,您可以使用@detail_route@list_route代替@link()。有关详细说明,请阅读"Routers", section "Extra link and actions""ViewSets", section "Marking extra actions for routing"的文档。