我为我的 API 公开的模型准备了一个标记系统。这些模型看起来像这样:
class TaggableModel(models.Model):
name = models.CharField(max_length=255)
tags = models.ManyToManyField(Tag, related_name="taggable_models")
class Tag(models.Model):
tag = models.CharField(max_length=32)
然后我设置了一个序列化程序并查看如下所示:
class TaggableModelSerializer(serializers.ModelSerializer):
class Meta:
model = TaggableModel
fields = ('id', 'name', 'tags',)
read_only_fields = ('id',)
class TaggableModelViewSet(viewsets.ModelViewSet):
queryset = TaggableModel.objects.all()
serializer_class = TaggableModelSerializer
permission_classes = (AllowAny,)
filter_backend = [DjangoFilterBackend]
filterset_fields = ['tags']
如果我想获取标签 ID 为 1、2 或 3 的所有 TaggableModels,我可以通过以下方式实现:
https://my-api-domain/api/taggable-models?tags=1&tags=2&tags=3
有没有一种方法可以在分隔符上进行拆分,这样我就可以将它们全部作为一个参数传递?例如:
https://my-api-domain/api/taggable-models?tags=1,2,3
看起来我可以编写自己的自定义 DjangoFilterBackend 过滤器,但我不确定从哪里开始。或者也许有更简单的方法来完成此操作?
请您参考如下方法:
当然,您可以通过具有特定字段“widget”的自定义过滤器集类来做到这一点(这就是它在 django-filters 中的调用方式)
这是您可以尝试的示例:
# filters.py
from django_filters.rest_framework import FilterSet, filters
from django_filters.widgets import CSVWidget
from .your_models import Tag, TaggableModel
class TaggableModelFilterSet(FilterSet):
tags = filters.ModelMultipleChoiceFilter(
queryset=Tag.objects.all(), widget=CSVWidget,
help_text=_("A list of ids, comma separated, identifying tags"),
method='filter_tags'
)
class Meta:
model = TaggableModel
fields = ['tags']
def filter_tags(self, queryset, name, value):
if value:
queryset = queryset.filter(tags__in=value)
return queryset
# views.py
class TaggableModelViewSet(viewsets.ModelViewSet):
queryset = TaggableModel.objects.all()
serializer_class = TaggableModelSerializer
permission_classes = (AllowAny,)
filter_backends = [DjangoFilterBackend]
filter_class = TaggableModelFilterSet
