我有两个模型,分别是 A 和 B。
模型看起来像
class A(models.Model):
name = models.CharField(max_length=100)
description = models.CharField(max_length=1000)
slug = models.CharField(max_length=100)
class B(models.Model):
name = models.CharField(max_length=100)
description = models.CharField(max_length=1000)
slug = models.CharField(max_length=100)
A = models.foreignkey(A, models.SET_NULL, blank=true)
序列化器
Class ASerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = A
fields = '__all__'
lookup_field = 'slug'
extra_kwargs = {'url': {'lookup_field': 'slug'}}
Class BSerializer(serializers.ModelSerializer):
class Meta:
model = B
fields = '__all__'
模型视图集
from rest_framework import viewsets, permissions
from rest_framework.response import Response
from rest_framework.decorators import action
from .models import (A, B)
from .serializers import (ASerializer, BSerializer)
Class AViewSet(viewsets.ModelViewSet):
queryset = A.objects.all()
serializer_class = ASerializer
permission_classes = (permissions.DjangoModelPermissionsOrAnonReadOnly, )
lookup_field = 'slug'
def retrieve(self, request, *args, **kwargs):
instance = self.get_object()
bs = B.objects.filter(A=instance.id)
serializer = BSerializer(bs, many=True)
return Response(serializer.data)
我想访问类似的网址
Url: ^A/<slug>/B/slug>/$
我浏览了 Django Restframework 文档,发现我们可以添加自定义 url,例如 ( drf-custom-rounting。我不知道如何在上面创建访问 url 模式。
怎么定制这样的?
编辑: 我已经解决了我的问题。我在这里找到了类似类型的问题解决方案。感谢您的所有回复。