How do I check for the existence of a process without a failed exit code being returned?

Question

I’m using Amazon Linux. I want to start a process provided one isn’t running already. This is a bash script I embed within a Jenkins Maven job. So I have

pids=$(pidof /usr/bin/Xvfb)
if [ ! -n "$pids" ]; then
    Xvfb :0 -screen 5 1024x768x8 &
fi

Unfortunately, if there is no process, the line “pids=$(pidof /usr/bin/Xvfb)” returns a failing exit code and none of the other lines following are executed. Is there a way I can write the above such that no failed exit codes will be returned?

Just use `if ! pidof /usr/bin/Xvfb; then : do something; fi` — cuonglm, Apr 28 '16 at 01:51
I tried "if ! pidof /usr/bin/Xvfb; then : Xvfb :0 -screen 5 1024x768x8 &; fi" and got the error, "syntax error near unexpected token `;'". — Dave, Apr 28 '16 at 02:24
Should be `if ! pidof /usr/bin/Xvfb; then Xvfb :0 -screen 5 1024x768x8 & fi"` if you want to run command in background. — cuonglm, Apr 28 '16 at 02:28

score 1 · Answer 1 · edited May 24 '20 at 06:41

If you're running under set -e, add || true after a command to ignore a failure of that command.

pids=$(pidof /usr/bin/Xvfb || true)
if [ ! -n "$pids" ]; then
    Xvfb :0 -screen 5 1024x768x8 &
fi

But since pidof returns a nonzero status if no processes are found, you could directly test its return status instead of checking whether its output is empty.

if ! pidof /usr/bin/Xvfb >/dev/null 2>&1; then
    Xvfb :0 -screen 5 1024x768x8 &
fi

How do I check for the existence of a process without a failed exit code being returned?

1 Answers1