Get name of current script in Python

Question

I\'m trying to get the name of the Python script that is currently running.

I have a script called foo.py and I\'d like to do something like this in ord

Answer 1

As of Python 3.5 you can simply do:

from pathlib import Path
Path(__file__).stem

See more here: https://docs.python.org/3.5/library/pathlib.html#pathlib.PurePath.stem

For example, I have a file under my user directory named test.py with this inside:

from pathlib import Path

print(Path(__file__).stem)
print(__file__)

running this outputs:

>>> python3.6 test.py
test
test.py

Answer 2

all that answers are great, but have some problems You might not see at the first glance.

lets define what we want - we want the name of the script that was executed, not the name of the current module - so __file__ will only work if it is used in the executed script, not in an imported module. sys.argv is also questionable - what if your program was called by pytest ? or pydoc runner ? or if it was called by uwsgi ?

and - there is a third method of getting the script name, I havent seen in the answers - You can inspect the stack.

Another problem is, that You (or some other program) can tamper around with sys.argv and __main__.__file__ - it might be present, it might be not. It might be valid, or not. At least You can check if the script (the desired result) exists !

the library lib_programname does exactly that :

• check if __main__ is present
• check if __main__.__file__ is present
• does give __main__.__file__ a valid result (does that script exist ?)
• if not: check sys.argv:
• is there pytest, docrunner, etc in the sys.argv ? --> if yes, ignore that
• can we get a valid result here ?
• if not: inspect the stack and get the result from there possibly
• if also the stack does not give a valid result, then throw an Exception.

by that way, my solution is working so far with setup.py test, uwsgi, pytest, pycharm pytest , pycharm docrunner (doctest), dreampie, eclipse

there is also a nice blog article about that problem from Dough Hellman, "Determining the Name of a Process from Python"

BTW, it will change again in python 3.9 : the file attribute of the main module became an absolute path, rather than a relative path. These paths now remain valid after the current directory is changed by os.chdir()

So I rather want to take care of one small module, instead of skimming my codebase if it should be changed somewere ...

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^{Disclaimer: I'm the author of the lib_programname library.}

Answer 3

The Above answers are good . But I found this method more efficient using above results.
This results in actual script file name not a path.

import sys    
import os    
file_name =  os.path.basename(sys.argv[0])

Answer 4

The first argument in sys will be the current file name so this will work

import sys
print sys.argv[0] # will print the file name

Answer 5

we can try this to get current script name without extension.

import os

script_name = os.path.splitext(os.path.basename(__file__))[0]

Answer 6

Since the OP asked for the name of the current script file I would prefer

import os
os.path.split(sys.argv[0])[1]