BASH: Bind Program's Existance to Another Program

Question

I want to create a script that runs two programs. That part's simple, but there's a catch: I want to kill the second one if the first one exits. How can I do that?

Edit

I tried killing the program by it's PID after the other one exits, but the program is hamster-time-tracker, which is a Python application that exits immediately, apparently spawning another process. How can I get around this? Is there some way to get the other PID spawned?

Edit 2

Figured it out. I had to run python /usr/bin/hamster-time-tracker instead of hamster-time-tracker, and it stayed running.

score 5 · Accepted Answer · answered May 07 '12 at 17:53

5

#!/bin/bash

cmd-a &
a=${!}

cmd-b &
b=${!}

wait $a
kill $b

I used yes a and yes b as commands when testing this.

answered May 07 '12 at 17:53

wfaulk

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+1 for an elegant solution that is easy to understand – Kenny Rasschaert May 07 '12 at 18:22

score 3 · Answer 2 · edited May 07 '12 at 18:21

3

There's wait command in the bash to wait for termination of first application and then kill second app.

Since waits are inserted automatically after commands not ending with &, the right order may save you from going into manual wait trouble:

A &
B          # waits
kill $!    # then kills A

edited May 07 '12 at 18:21

yrk

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answered May 07 '12 at 17:45

OJ278

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you can find the pid of spawned process using ps aux | grep and then wait for it. – OJ278 May 07 '12 at 18:05
That's not robust in the case of running multiple copies of the same program. – nickgrim May 07 '12 at 19:41

BASH: Bind Program's Existance to Another Program

2 Answers2