Linux – Bash script that starts a process for 4 seconds, then kills it doesn’t work

bashcaptive-portallinux

I have a script that starts firefox for 4 seconds, then kills it. Firefox will automatically log into a captive portal, so I only need it to be open for 4 seconds, as soon as the wifi connects. I am on Ubuntu 13.04.

My problem seems to be that $pid isn't set.

firefox ; pid=$!
sleep 4
kill $pid

EDIT: removed set, and now it gives kill an invalid pid.

Best Answer

Your script does not work, because it waits until the firefox process has ended and afterwards executes pid=$!and the other command.

An easy way to do what you want is timeout:

timeout 4s firefox

It starts the program provided after the first argument and stops it after the time given as the first argument has passed.

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Linux – Process not Listed by PS or in /proc/

The $! PID is that of su since you're backgrounding it rather than your $_CMD. You should move the & inside the closing quote.

Use pidof or pgrep to get your process' PID instead of using $!. If you're on a system that uses Upstart or has a similar daemon system, you should model your script after those and use the functions in /lib/lsb/init-functions or similar.

Linux – Shell Script to Start and Kill Processes Sequentially

You can use bash job control:

#!/bin/bash

./server &
./client
kill %1

Be sure to put the #!/bin/bash at the beginning of the script so that bash is used to execute the script (I'm not sure if job control is supported in sh, please correct me if it does).

Best Answer

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