Terminate python code without letting it finish if while loop condition is unmet in bash

Question

I have this so far. The issue is, if pythoncode.py started before 8*60*60 seconds, then it can still continue running beyond 8*60*60 seconds until it finishes and only then will it stop. I need it to terminate pythoncode.py at exactly (loosely, +- 1 second is not an issue) 8*60*60 seconds even if it is running.

#! /bin/bash
end=$((SECONDS+8*60*60))
while [ $SECONDS -lt $end ]; do
    python3 /Users/Name/Desktop/pythoncode.py
done

I'm a beginner so explaining what to do specifically with this example would be appreciated over just a general answer. Thanks and happy new year.

Possibly related: Automatically kill process if its runtime exceeds some predetermined value — steeldriver, Jan 02 '21 at 02:15

score 1 · Accepted Answer · answered Jan 02 '21 at 01:19

1

You can call your python script in the background and use sleep function as a timer :

#Execute the script in the background
python3 /Users/Name/Desktop/pythoncode.py & 
#Get its process id
pypid=$!
#Wait for the needed period in seconds
sleep "$end"
#Then force the process to be terminated
kill -9 $pypid

answered Jan 02 '21 at 01:19

Reda Salih

1,754

2

Would this accomplish the exact same thing? timeout 8h sh -c 'while true; do python3 /Users/Name/Desktop/pythoncode.py; done' – Ardalan Ghazizadeh Jan 02 '21 at 04:31
Yes same thing here if you do have timeout installed :) – Reda Salih Jan 02 '21 at 14:27
/usr/bin/timeout is in the coreutils package, so most probably it's installed. – glenn jackman Jan 02 '21 at 15:18

Terminate python code without letting it finish if while loop condition is unmet in bash

1 Answers1