Spawn a new non-blocking process using Python on Mac OS X

问题

I found some articles and even stack|overflow questions addressing this subject, but I still can't do it..

What I want to do is open an instance of firefox from python. then the python application should keep minding its own business and ignore the firefox process.

I was able to achive this goal on Windows-7 and XP using:

subprocess.Popen()

On OS X I tried:

subprocess.Popen(['/Applications/Firefox.app/Contents/MacOS/firefox-bin'])
subprocess.call(['/Applications/Firefox.app/Contents/MacOS/firefox-bin'])
subprocess.call(['/Applications/Firefox.app/Contents/MacOS/firefox-bin'], shell=True)
os.system('/Applications/Firefox.app/Contents/MacOS/firefox-bin')

(and probably some others I forgot) to no avail. My python app freezes till I go and close the firefox app.

What am I missing here? any clues?

回答1:

You'll need to detach the process somehow. I snatched this from spawning process from python

import os
pid = os.fork()
if 0 == pid:
  os.system('firefox')
  os._exit(0)
else:
  os._exit(0)

This spawns a forked headless version of the same script which can then execute whatever you like and quit directly afterwards.

回答2:

To show what I meant:

import os
if not os.fork():
    os.system('firefox')
os._exit(0)

Version that doesn't quit the main Python process:

import os
if not os.fork():
    os.system('firefox')
    os._exit(0)

来源：https://stackoverflow.com/questions/6441807/spawn-a-new-non-blocking-process-using-python-on-mac-os-x

标签

python

macos

subprocess

nonblocking