Little Bash Script: Catch Errors?

Question

I've written (well, remixed to arrive at) this Bash script

# pkill.sh
trap onexit 1 2 3 15 ERR

function onexit() {
    local exit_status=${1:-$?}
    echo Problem killing $kill_this
    exit $exit_status
}

export kill_this=$1
for X in `ps acx | grep -i $1 | awk {'print $1'}`; do
  kill $X;
done

it works fine but any errors are shown to the display. I only want the echo Problem killing... to show in case of error. How can I "catch" (hide) the error when executing the kill statement?

Disclaimer: Sorry for the long example, but when I make them shorter I inevitably have to explain "what I'm trying to do."

Answer 1

You can redirect stderr and stdout to /dev/null via something like pkill.sh > /dev/null 2>&1. If you only want to suppress the output from the kill command, only apply it to that line, e.g., kill $X > /dev/null 2>&1;

What this does is take send the standard output (stdout) from kill $X to /dev/null (that's the > /dev/null), and additionally send stderr (the 2) into stdout (the 1).

Answer 2

# pkill.sh
trap onexit 1 2 3 15 ERR

function onexit() {
    local exit_status=${1:-$?}
    echo Problem killing $kill_this
    exit $exit_status
}

export kill_this=$1
for X in `ps acx | grep -i $1 | awk {'print $1'}`; do
    kill $X 2>/dev/null
    if [ $? -ne 0 ]
    then
        onexit $?
    fi
done

Answer 3

For my own notes, here's my new code using Paul Creasey's answer:

# pkill.sh: this is dangerous and should not be run as root!
trap onexit 1 2 3 15 ERR

#--- onexit() -----------------------------------------------------
#  @param $1 integer  (optional) Exit status.  If not set, use `$?'
function onexit() {
    local exit_status=${1:-$?}
    echo Problem killing $kill_this
    exit $exit_status
}

export kill_this=$1
for X in `ps acx | grep -i "$1" | awk {'print $1'}`; do
  kill $X 2>/dev/null
done

Thanks all!