Django - 模型类接口？(Django - Model class interfaces?)

 我有一些与各种Web API交互的模型。 我需要确保每个模型都有一个特定的方法。 使用PHP我会创建一个类接口，以确保我的模型具有完成它的工作所必需的方法，但似乎Python接口不适用于Django模型。 
 
 我假设这样做的方法是创建一个扩展model.Model的基类，它定义了我需要的方法，如果需要，我可以在每个API模型中覆盖它们。 如果没有Django在同步数据库时拿起“基础”类，我怎么能做到这一点？ 这甚至是正确的方法吗？ 

I have a selection of models which interact with various web APIs. I need to ensure each of the models has a particular method. With PHP I'd create a class interface to ensure my model has the methods necessary to do it's job but it seems Python interfaces don't work with Django models. 
 
I'm assuming the way to do this would be to create a base class that extends model.Model which defines the methods I need and I can overwrite them in each API model if necessary. How am I able to do this without Django picking up the "base" class when syncing the database? Is this even the right way to do it?

原文：https://stackoverflow.com/questions/6033817