Is there an platform independent equivalent of os.startfile()?

Question

I want to run a program on several platforms (including Mac OS), so I try to keep it as platform independent as possible. I use Windows myself, and I have a line os.st

Answer 1

It appears that a cross-platform file opening module does not yet exist, but you can rely on existing infrastructure of the popular systems. This snippet covers Windows, MacOS and Unix-like systems (Linux, FreeBSD, Solaris...):

import os, sys, subprocess

def open_file(filename):
    if sys.platform == "win32":
        os.startfile(filename)
    else:
        opener = "open" if sys.platform == "darwin" else "xdg-open"
        subprocess.call([opener, filename])