vscode 从 python 脚本中获取当前文件的路径

Question

When ran from a terminal, sys.argv[0] is the path of current script, but in python interactive that variable points to "/some/path/ipykernel_launcher.py" which is a temporary file.

How do I get the path of current script (which I am editing in vscode)? I need this information because whenever I create a file, I automatically log which script created it. For that, I overload the open() function to automatically log the creation. But when file is created from a python interactive session, I am missing such information.

Answer 1

import os

print(os.getcwd())

I think that's what you want. It prints the current working directory.

Answer 2

The answer that works for me, obtained from How do I get the path and name of the file that is currently executing? is this:

os.path.realpath(__file__)

I tried all other suggestions, but didn't work.