如何从执行但未声明的函数中获取模块的文件路径? [英] How to get the file path of a module from a function executed but not declared in it?

问题描述

如果我想要当前模块的路径，我将使用__file__.

If I want the path of the current module, I'll use __file__.

现在让我们说我想要一个函数将其返回.我做不到:

Now let's say I want a function to return that. I can't do:

def get_path():
    return __file__

因为它将返回已在其中声明该函数的模块的路径.

Because it will return the path of the module the function has been declared in.

即使函数不是在模块的根目录而是在任何嵌套级别调用，我也需要它工作.

I need it to work even if the function is not called at the root of the module but at any level of nesting.

推荐答案

这就是我要这样做的方式:

This is how I would do it:

import sys

def get_path():
    namespace = sys._getframe(1).f_globals  # caller's globals
    return namespace.get('__file__')

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