How to spawn a new independent process in Python

python subprocess fork daemon spawn
Marc picture Marc · Dec 18, 2013 · Viewed 39.2k times · Source

I have a some Python code that occasionally needs to span a new process to run a shell script in a "fire and forget" manner, i.e. without blocking. The shell script will not communicate with the original Python code and will in fact probably terminate the calling Python process, so the launched shell script cannot be a child process of the calling Python process. I need it to be launched as an independent process.

In other words, let's say I have mycode.py and that launches script.sh. Then mycode.py will continue processing without blocking. The script script.sh will do some things independently and will then actually stop and restart mycode.py. So the process that runs script.py must be completely independent of mycode.py. How exactly can I do this? I think subprocess.Popen will not block, but will still create a child process that terminates as soon as mycode.py stops, which is not what I want.

Answer

Lasse picture Lasse · Dec 18, 2013

Try prepending "nohup" to script.sh. You'll probably need to decide what to do with stdout and stderr; I just drop it in the example.

import os
from subprocess import Popen

devnull = open(os.devnull, 'wb') # Use this in Python < 3.3
# Python >= 3.3 has subprocess.DEVNULL
Popen(['nohup', 'script.sh'], stdout=devnull, stderr=devnull)

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