在堆栈中上一级获取函数的__file__ [英] Get the __file__ of the function one level up in the stack

问题描述

我发现我经常使用这种模式:

I've found that I'm using this pattern a lot :

os.path.join(os.path.dirname(__file__), file_path)

所以我决定在一个包含许多小实用程序的文件中放入一个函数:

so I've decided to put in a function in a file that has many such small utilities:

def filepath_in_cwd(file_path): 
    return os.path.join(os.path.dirname(__file__), file_path)

问题是，__file__返回 current 文件，因此返回当前文件夹，而我漏掉了所有要点.我可以进行这种丑陋的修改(或者只是继续按原样编写模式):

The thing is, __file__ returns the current file and therefore the current folder, and I've missed the whole point. I could do this ugly hack (or just keep writing the pattern as is):

def filepath_in_cwd(py_file_name, file_path): 
    return os.path.join(os.path.dirname(py_file_name), file_path)

，然后对其的调用将如下所示:

and then the call to it will look like this:

filepath_in_cwd(__file__, "my_file.txt")

，但是如果我有办法将堆栈中的函数的__file__向上一级，我会更喜欢它.有什么办法吗?

but I'd prefer it if I had a way of getting the __file__ of the function that's one level up in the stack. Is there any way of doing this?

推荐答案

这应该做到:

inspect.getfile(sys._getframe(1))

sys._getframe(1) 获取调用者框架， inspect.getfile(...) 检索文件名.

sys._getframe(1) gets the caller frame, inspect.getfile(...) retrieves the filename.

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