你如何使用带有参数列表的 django-filter 包?
How do you use the django-filter package with a list of parameters?
我想用 django-filter 过滤我的模型。如果我按一个 id 过滤,它工作正常:
http://localhost:8000/accommodations?accommodationType_id=1
但我不知道如何按多个 ID 进行过滤,例如。
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')
希望有解决办法
我找到了以下解决方案来解决我的问题:)
https://gist.github.com/aBuder/654fb945f085b17358d8
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
我知道这是一个老问题,但可能值得提供更新的答案。
Django-filter 贡献者添加了一个名为 BaseInFilter
的字段,您可以将其与其他过滤器结合使用以验证内容。
查看文档:
https://django-filter.readthedocs.io/en/latest/ref/filters.html#baseinfilter
例如,这适用于您的情况:
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', ]
然后您将能够按 ID 列表进行过滤:http://localhost:8000/accommodations?accommodationType_id_in=1,2
现在有一种更简单的方法可以实现这一点。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,您可以在提供参数列表之前将 __in
添加到过滤器,它会按您预期的那样工作:
http://localhost:8000/accommodations?accommodationType_id__in=1,2
关于查找过滤器可用的 django-filter
文档很差,但是 in
查找过滤器在 Django documentation 本身中提到。
我想用 django-filter 过滤我的模型。如果我按一个 id 过滤,它工作正常:
http://localhost:8000/accommodations?accommodationType_id=1
但我不知道如何按多个 ID 进行过滤,例如。
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')
希望有解决办法
我找到了以下解决方案来解决我的问题:)
https://gist.github.com/aBuder/654fb945f085b17358d8
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
我知道这是一个老问题,但可能值得提供更新的答案。
Django-filter 贡献者添加了一个名为 BaseInFilter
的字段,您可以将其与其他过滤器结合使用以验证内容。
查看文档: https://django-filter.readthedocs.io/en/latest/ref/filters.html#baseinfilter
例如,这适用于您的情况:
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', ]
然后您将能够按 ID 列表进行过滤:http://localhost:8000/accommodations?accommodationType_id_in=1,2
现在有一种更简单的方法可以实现这一点。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,您可以在提供参数列表之前将 __in
添加到过滤器,它会按您预期的那样工作:
http://localhost:8000/accommodations?accommodationType_id__in=1,2
关于查找过滤器可用的 django-filter
文档很差,但是 in
查找过滤器在 Django documentation 本身中提到。