Get Python function's owning class from decorator

问题

I have a decorator in PY. It is a method and takes the function as a parameter. I want to create a directory structure based based on the passed function. I am using the module name for the parent directory but would like to use the classname for a subdirectory. I can't figure out how to get the name of the class that owns the fn object.

My Decorator:

def specialTest(fn):
    filename = fn.__name__
    directory = fn.__module__
    subdirectory = fn.__class__.__name__ #WHERE DO I GET THIS

回答1:

If fn is an instancemethod, then you can use fn.im_class.

>>> class Foo(object):
...     def bar(self):
...         pass
...
>>> Foo.bar.im_class
__main__.Foo

Note that this will not work from a decorator, because a function is only transformed into an instance method after the class is defined (ie, if @specialTest was used to decorate bar, it would not work; if it's even possible, doing it at that point would have to be done by inspecting the call stack or something equally unhappy).

回答2:

In Python 2 you can use im_class attribute on the method object. In Python 3, it'll be __self__.__class__ (or type(method.__self__)).

来源：https://stackoverflow.com/questions/7680446/get-python-functions-owning-class-from-decorator