Signalling a bash script running in the background in an infinite loop

Question

I have a bash script (sleeping in an infinite loop) running as a background process. I want to periodically signal this process, to do some processing without killing the script, ie. the script on receiving the signal should run a function and then go back to sleep. How do I signal the background process without killing it?

Here's what the code looks like for the script test.sh:

MY_PID=$$
echo $MY_PID > test.pid

while true
do
  sleep 5
done
trap 'run' SIGUSR1 

run()
{
 // data processing
}

This is how I am running it and triggering it:

run.sh &

kill -SIGUSR1 `cat test.pid`

First of all, you have to move the trap command above the while loop, or else it will never be executed. And the run function as well for that matter. — glenn jackman
– glenn jackman, Commented Jan 18, 2014 at 15:57

Community · Accepted Answer · 2017-05-23 11:43:27Z

3

This works for me.

EDITED 2014-01-20: This edit avoids the frequent waking up. Also see this question: Bash: a sleep in a while loop gets its own pid

#!/bin/bash

MY_PID=$$
echo $MY_PID > test.pid

trap 'kill ${!}; run' SIGUSR1 

run()
{
    echo "signal!"
}

while true
do
  sleep 1000  & wait ${!}
done

Running:

> ./test.sh &
[1] 14094
> kill -SIGUSR1 `cat test.pid`
> signal!
>

edited May 23, 2017 at 11:43

CommunityBot

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answered Jan 18, 2014 at 18:06

Christian Fritz

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2 Comments

egorulz Over a year ago

Is there some way to do this without waking my process up every other second. I don't know if bash provides this option, but does tcl or some other scripting language provide it i wonder. Wherein it just sleeps until signal.

egorulz Over a year ago

Awesome, works like a charm. I also added in some code to kill the sleep silently. trap 'killsub; parse' SIGUSR1 function killsub() { kill -9 ${!} 2>/dev/null wait ${!} 2>/dev/null }

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