How to stop a running bash script after it's current running command finished?

Question

There is a shell script:

#!/bin/bash
python3 do_something_1.py
python3 do_something_2.py
python3 do_something_3.py
python3 do_something_4.py
python3 do_something_5.py

If it's running python3 do_something_3.py now.

How can I tell bash process that after python3 do_something_3.py exits by itself(don't kill subprocess, wait it), no longer run subsequent commands python3 do_something_4.py,python3 do_something_5.py anymore?

Answer 1

You can use a while loop to check if do_something_3.py is still running (e.g. using pgrep) and sleep as long it is.

After the loop, kill your script.

while pgrep -f do_something_3.py >/dev/null ; then
    sleep 0.1
done && pkill -9 myscript

• The exact pkill command depends how you named and called the script, you might want to test it with pgrep myscript or find out the PID and use kill -9 PID.
• Note, that this is nowhere ideal, because you won't find the exact timing to stop your script before do_something_4.py has started.

Answer 2

Signals are the simplest solution here. From the Bash man page:

If bash is waiting for a command to complete and receives a signal for which a trap has been set, the trap will not be executed until the command completes.

An example (thescript.sh):

trap 'echo "Exiting on USR1 after command"; exit' USR1

echo 'Beginning'
sleep 2  # long-running command
echo 'Middle'
sleep 2  # long-running command
echo 'End'

This will exit after the first sleep completes:

./thescript.sh &
pid="$!"
sleep 1
kill -USR1 "$pid"  # roughly the middle of first sleep
wait -n "$pid"

This will exit after the second sleep completes:

./thescript.sh &
pid="$!"
sleep 3
kill -USR1 "$pid"  # roughly the middle of second sleep
wait -n "$pid"