【发布时间】:2015-09-22 12:17:25
【问题描述】:
我有一个简单的问题:目前我可以这样做以从后端获取对象:
http://127.0.0.1:8000/api/v1/boats/boats?id=10
http://127.0.0.1:8000/api/v1/boats/boats?home_port=98&id=5
但我想根据 id 列表或 home_ports 列表获取一组船,我已经尝试过:
http://127.0.0.1:8000/api/v1/boats/boats?id=10,11
http://127.0.0.1:8000/api/v1/boats/boats?id_in=10,11
http://127.0.0.1:8000/api/v1/boats/boats?id=10,id=11
http://127.0.0.1:8000/api/v1/boats/boats?id=10&id=11
但是这些不起作用。使用 django-filter 执行此操作的最佳方法是什么,URL 规则是如何定义的?
这是我的看法:
class BoatList(generics.ListCreateAPIView):
permission_classes = (IsOwnerOrReadOnly,)
serializer_class = BoatSerializer
queryset = Boat.objects.all()
filter_backends = (filters.DjangoFilterBackend,)
filter_fields = ('id', 'home_port',)
我标记为接受的解决方案 100% 回答了我的问题,但我最终根据我发现的另一个使用过滤器的帖子实施了一些不同的解决方案:
class ListFilter(Filter):
def filter(self, qs, value):
if not value:
return qs
self.lookup_type = 'in'
values = value.split(',')
return super(ListFilter, self).filter(qs, values)
class BoatFilter(FilterSet):
ids = ListFilter(name='id')
class Meta:
model = Boat
fields = ['home_port', 'ids']
class BoatList(generics.ListCreateAPIView):
permission_classes = (IsOwnerOrReadOnly,)
serializer_class = BoatSerializer
queryset = Boat.objects.all()
filter_backends = (filters.DjangoFilterBackend,)
filter_class = BoatFilter
def perform_create(self, serializer):
serializer.save(owner=self.request.user)
【问题讨论】:
标签: python django api django-rest-framework django-filter