Python 从Django/DRF中的公共视图中提取特定模型的标准方法？

我目前有两个基本相同的模型（从基础模型继承），但从通用视图引用它们时遇到问题：

型号：

class BaseModel(models.Model):
    name = models.CharField(...)
    owner = ForeignKey(...)

class Cat(BaseModel):
    ...

class Dog(BaseModel):
    ...

class CommonViewset(viewsets.ModelViewSet):
    @link()
    def set_owner(self, request, pk=None):

        #how do I get Cat or Dog models cleanly here?

        #super fugly/unstable way
        url_path = request.META['PATH_INFO']
        if 'cats' in url_path:
            Cat.objects.get(pk=pk).owner = ...
        elif 'dogs' in url_path:
            Dog.objects.get(pk=pk).owner = ...

查看：

class BaseModel(models.Model):
    name = models.CharField(...)
    owner = ForeignKey(...)

class Cat(BaseModel):
    ...

class Dog(BaseModel):
    ...

class CommonViewset(viewsets.ModelViewSet):
    @link()
    def set_owner(self, request, pk=None):

        #how do I get Cat or Dog models cleanly here?

        #super fugly/unstable way
        url_path = request.META['PATH_INFO']
        if 'cats' in url_path:
            Cat.objects.get(pk=pk).owner = ...
        elif 'dogs' in url_path:
            Dog.objects.get(pk=pk).owner = ...

---

我也可以将

set\u owner

链接放在不同的视图中，但这感觉很不干燥。提前感谢您的关注

您可以传递模型以在类的

as\u view

方法中使用：

url(r'^cats/my-url/$', CommonViewSet.as_view(model=Cat)),

---

ModelViewSet

类继承自Django的

View

类，因此这将在视图集实例上设置

model

属性。然后可以使用

self.model

为当前url获取正确的模型

谢谢！我正在使用路由器，因此模型将设置在视图集中。