从装饰器获取Python函数的拥有类 [英] Get Python function's owning class from decorator

问题描述

我在PY有个装饰师。它是一个方法，并将该函数作为参数。我想基于传递的函数创建一个目录结构。我使用模块名称为父目录，但想使用类名为子目录。我不知道如何获得拥有fn对象的类的名称。

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.

我的装饰：

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

推荐答案

如果 fn 是实例方法，那么您可以使用 fn.im_class 。

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

请注意， em> 从装饰器工作，因为一个函数只是在之后被转换为一个实例方法（即 @specialTest 用于装饰 bar ，它不会工作;如果甚至可能，在这一点上做它将必须通过检查调用堆栈或一些同等不愉快）。

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).

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