我想用django-filter过滤我的模型。如果我按一个id过滤它可以正常工作:
http://localhost:8000/accommodations?accommodationType_id=1
但我不知道如何通过多个ID进行过滤。
http://localhost:8000/accommodations?accommodationType_id=1,2
我的实际ViewSet
看起来像这样:
class AccommodationViewSet(viewsets.ReadOnlyModelViewSet):
"""
REST API endpoint for 'accommodation' resource
"""
queryset = Accommodation.objects.all()
serializer_class = AccommodationSerializer
filter_backends = (filters.DjangoFilterBackend,)
filter_fields = ('accommodationType_id', 'name')
我希望有一个解决方案。
答案 0 :(得分:13)
我为我的问题找到了以下解决方案:)
from webapp.serializers import *
from rest_framework import viewsets
from rest_framework import filters
from django_filters import Filter, FilterSet
class ListFilter(Filter):
def filter(self, qs, value):
if not value:
return qs
# For django-filter versions < 0.13, use lookup_type instead of lookup_expr
self.lookup_expr = 'in'
values = value.split(',')
return super(ListFilter, self).filter(qs, values)
class AccommodationFilter(FilterSet):
ids = ListFilter(name='id')
accommodationType_ids = ListFilter(name='accommodationType_id')
class Meta:
model = Accommodation
fields = ['ids', 'accommodationType_ids']
class AccommodationViewSet(viewsets.ReadOnlyModelViewSet):
"""
REST API endpoint for 'accommodation' resource
"""
queryset = Accommodation.objects.all()
serializer_class = AccommodationSerializer
filter_backends = (filters.DjangoFilterBackend,)
filter_class = AccommodationFilter
答案 1 :(得分:1)
我知道这是一个古老的问题,但值得给出更新的答案。
Django过滤器提供者添加了一个名为BaseInFilter
的字段,您可以将其与其他过滤器组合以验证内容。
查看文档: https://django-filter.readthedocs.io/en/latest/ref/filters.html#baseinfilter
例如,这将适合您的情况:
from django_filters import rest_framework as filters
class NumberInFilter(filters.BaseInFilter, filters.NumberFilter):
pass
class AccommodationFilter(filters.FilterSet):
accommodationType_id_in = NumberInFilter(field_name='accommodationType_id', lookup_expr='in')
class Meta:
model = Accommodation
fields = ['accommodationType_id_in', ]
然后,您将可以按ID列表进行过滤:http://localhost:8000/accommodations?accommodationType_id_in=1,2
答案 2 :(得分:0)
现在有一种更简单的方法可以实现这一目标。在 django-filter
documentation 的深处,它提到您可以使用“映射到查找列表的字段名称字典”。
您的代码将像这样更新:
class AccommodationViewSet(viewsets.ReadOnlyModelViewSet):
"""
REST API endpoint for 'accommodation' resource
"""
queryset = Accommodation.objects.all()
serializer_class = AccommodationSerializer
filter_backends = (filters.DjangoFilterBackend,)
filter_fields = {
'accommodationType_id': ["in", "exact"], # note the 'in' field
'name': ["exact"]
}
现在在 URL 中,您需要在提供参数列表之前将 __in
添加到过滤器中,它会按您的预期工作:
http://localhost:8000/accommodations?accommodationType_id__in=1,2
关于可用查找过滤器的 django-filter
文档非常糟糕,但 Django documentation 本身中提到了 in
查找过滤器。