Django DRF - 添加 subview /url？

Question

我想在更改查询集的 View 下创建一个 subview (如果这是正确的术语？)

parent URL
mysite.com/api/sites

Child URL
mystic.com/apit/sites/open

而且每个 URL 都可以这样搜索

parent URL
mysite.com/api/sites/search=London

Child URL
mystic.com/api/sites/open/search=London

我的父 View 、序列化程序和 URL 已存在

class SiteROView(viewsets.ReadOnlyModelViewSet):
    queryset = Site.objects.all()
    serializer_class = SiteSerializer
    permission_classes = (IsAdminUser,)
    filter_class = Site
    filter_backends = (filters.SearchFilter,)
    search_fields = ('location','postcode','state')

所以我想我需要以某种方式添加子网址

class SiteROView(viewsets.ReadOnlyModelViewSet):
    queryset = Site.objects.all()
    serializer_class = SiteSerializer
    permission_classes = (IsAdminUser,)
    filter_class = Site
    filter_backends = (filters.SearchFilter,)
    search_fields = ('location','postcode','state')

    def url_open:
        queryset = Site.objects.filter(state='open')

这可能吗？我该如何实现？

谢谢

Answer 1

您可以使用detail_route装饰器来做到这一点

from rest_framework.response import Response

class SiteROView(viewsets.ReadOnlyModelViewSet):
    ..........
    # your codes up here

    @list_route(methods=['get'],url_path='open' permission_classes=[YourPermissionClass])
    def open(self, request, *args, **kwargs):
         # your rest of code and response
         queryset = <your_filtered_queryset>
         serializer = self.serializer_class(queryset, many=True)
         return Response(data=serializer.data)