获取堆栈中上一级函数的 __file__

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这样的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，我错过了重点.我可以做这个丑陋的 hack(或者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)

然后对它的调用将如下所示:

and then the call to it will look like this:

filepath_in_cwd(__file__, "my_file.txt")

但如果我有办法获取堆栈中上一层的函数的 __file__ ，我会更喜欢它.有没有办法做到这一点?

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

应该这样做:

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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