Python 是否覆盖Django模型的副本？

复制Django模型对象时，我总是希望它与我从中复制的源对象不具有相同的标识（即主键）。通常你会这样做：

copy = copy.copy(source)
copy.pk = None
copy.save()

class MyModel(Model):
    def __copy__(self):
        result = MyModel.__new__(MyModel)
        result.__dict__.update(self.__dict__)
        result.pk = None
        return result

但是我想让它自动的像这样：

copy = copy.copy(source)
copy.pk = None
copy.save()

class MyModel(Model):
    def __copy__(self):
        result = MyModel.__new__(MyModel)
        result.__dict__.update(self.__dict__)
        result.pk = None
        return result

所以我现在可以做：

copy.copy(my_Model).save()

到目前为止，这似乎是可行的。但是，看看Django在

\uuuu init\uuuuu

和

\uuuu new\uuuuuu

中对模型对象所做的操作，我一直在想这是否确实是允许的，是否符合Django原则。令人惊讶的是，Django模型似乎没有覆盖

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。所以它不会与此冲突

此外，
\uuu eq\uu
方法也通过
pk
进行比较，因此这也不重要

长话短说，没有什么能阻止您重写
\uuu copy\uu
，只要您确保
pk
值在过程中发生更改。
谢谢matsjoyce，我不知道。