How do I get the path and name of the file that is currently executing

Question

I have scripts calling other script files but I need to get the filepath of the file that is currently running within the process.

For example, let's say I have three files. Using execfile:

• script_1.py calls script_2.py.
• In turn, script_2.py calls script_3.py.

How can I get the file name and path of script_3.py, from code within script_3.py, without having to pass that information as arguments from script_2.py?

Answer 1

Hello @kartik,

Since Python 3 is fairly mainstream, I wanted to include a pathlib answer, as I believe that it is probably now a better tool for accessing file and path information.

from pathlib import Path

current_file: Path = Path(__file__).resolve()

If you are seeking the directory of the current file, it is as easy as adding .parent to the Path() statement:

current_path: Path = Path(__file__).parent.resolve()

Thank You!!