Python Django Rest框架：如何实现嵌套逻辑？

假设我有三个模型：

class User(AppModel):
    name = models.CharField(max_length=255)

class Business(AppModel):
    owner = models.ForeignKey("User", related_name="businesses", on_delete=models.CASCADE)
    legal_name = models.CharField(max_length=255)

class Invoice(AppModel):
    business = models.ForeignKey("Business", related_name="invoices", on_delete=models.CASCADE)
    amount = models.integerField()

如您所见，

用户

可以有多个

业务

，而

业务

可以有多个

发票

My serializers.py：

class UserSerializer(serializers.ModelSerializer):
    class Meta:
        model = User
        fields= ('name')

class BusinessSerializer(serializers.ModelSerializer):
    owner = UserSerializer(many=False)
    class Meta:
        model = Business
        fields= ('owner','legal_name')

class InvoiceSerializer(serializers.ModelSerializer):
    business= BusinessSerializer(many=False)
    class Meta:
        model = Invoice
        fields= ('business','amount')

views.py：

class UserViewSet(viewsets.ModelViewSet):
    queryset = User.objects.all()
    serializer_class = UserSerializer

class BusinessViewSet(viewsets.ModelViewSet):
    queryset = Business.objects.all()
    serializer_class = BusinessSerializer

class InvoiceViewSet(viewsets.ModelViewSet):
    queryset = Invoice.objects.all()
    serializer_class = InvoiceSerializer

URL.py：

router = DefaultRouter()
router.register('user', UserViewSet, base_name='users')
router.register('business', BusinessViewSet, base_name='businesses')
router.register('invoice', InvoiceViewSet, base_name='invoices')
urlpatterns = router.urls

http://example.com/api/user

返回所有用户。没问题

但我想要的功能是：

• http://example.com/api/business/

  返回

  [
  {
  “法定名称”：“1业务”，
  “所有者”：1，
  },
  {
  “法定名称”：“2业务”，
  “所有者”：1，
  }，]

• http://example.com/api/business/1/

  返回

  {
  “法定名称”：“1业务”，
  “所有者”：1，
  }

以上是可以的。但我还需要：

• http://example.com/api/business/1/invoices/

  应该返回

  [
  {
  "业务":1,，
  “金额”：100，
  },
  {
  "业务":1,，
  “金额”：999，
  }，]

此外，我应该能够创建更新删除这些发票那里

---

有什么帮助吗？我是django rest框架的新手。上面的类只是一个示例。忽略错误。

您应该使用django装饰器，它们是视图集的

@list\u route

和

@detail\u route

。但是要小心你的DRF版本。因为这些装饰器在DRF 3.8+中合并为

@action

。这是你的电话号码

然后，您将能够从调用此端点

http://example.com/api/business/{{PK}}/invoices/
http://example.com/api/business/1/invoices/
http://example.com/api/business/3/invoices/
http://example.com/api/business/23/invoices/

在这里，您可以从中找到有关

@操作的更多详细信息

PS：别忘了在代码中控制空实体结果。您应该返回带有正确状态代码的正确响应

http://example.com/api/business/{{PK}}/invoices/
http://example.com/api/business/1/invoices/
http://example.com/api/business/3/invoices/
http://example.com/api/business/23/invoices/