我有一个简单的问题:目前我可以这样做从后端获取一个对象:
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
但是我想根据一系列ID或者home_ports列表获得一系列船只,我试过了:
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
但这些都行不通。使用django-filter执行此操作的最佳方法是什么,如何定义URL规则?
以下是我的观点:
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现了不同的内容:
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)
答案 0 :(得分:5)
doniyor给出的答案非常贴切。但我猜request
在使用它时无法使用。
还有另一种方法。您可以覆盖get_queryset
方法。这可以按如下方式完成:
class BoatList(generics.ListCreateAPIView):
permission_classes = (IsOwnerOrReadOnly,)
serializer_class = BoatSerializer
filter_backends = (filters.DjangoFilterBackend,)
filter_fields = ('id', 'home_port',)
def get_queryset(self):
id_list = self.request.GET.getlist("id")
if not id_list:
return []
return Boat.objects.filter(id__in=id_list)
答案 1 :(得分:3)
试试这个:
url:http://127.0.0.1:8000/api/v1/boats/boats?id=10,11
class BoatList(generics.ListCreateAPIView):
permission_classes = (IsOwnerOrReadOnly,)
serializer_class = BoatSerializer
queryset = Boat.objects.filter(id__in=request.GET.getlist('id')) #<------
filter_backends = (filters.DjangoFilterBackend,)
filter_fields = ('id', 'home_port',)
答案 2 :(得分:1)
您只需创建一个过滤器类
class NumberInFilter(BaseInFilter, NumberFilter):
pass
class myFilter(FilterSet):
id__in = NumberInFilter(field_name='id', lookup_expr='in')
class Meta:
model = Boat
在视图集中使用
filter_class = myFilter
您可以看到django-filter的此文档: https://django-filter.readthedocs.io/en/master/ref/filters.html?highlight=BaseInFilter