vscode get path of current file from inside python script

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.

Use the __file__ variable or the inspect module. Already answered here. — koorkevani
– koorkevani, Commented May 20, 2021 at 18:59
Thanks @koorkevani! The inspect strategy in that answer doesn't work, but __file__ does! — agarrubio
– agarrubio, Commented May 24, 2021 at 6:46

Ethitlan · Accepted Answer · 2021-05-20 19:06:31Z

1

import os

print(os.getcwd())

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

answered May 20, 2021 at 19:06

Ethitlan

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agarrubio Over a year ago

Sorry, but that gives the execution path, which, by the way, I usually set with os.chdir(some_dir) . It is true that at start of the interactive session, the os.getcwd() is that of the script. ... Anyway, I am missing the filename.

agarrubio · Accepted Answer · 2021-05-24 06:50:04Z

0

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.

answered May 24, 2021 at 6:50

agarrubio

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