Get exit code for command in bash/ksh

Question

I want to write code like this:

command="some command"  safeRunCommand $command  safeRunCommand() {    cmnd=$1     $($cmnd)     if [ $? != 0 ]; then       printf "Error when executing command: '$command'"       exit $ERROR_CODE    fi } 

But this code does not work the way I want. Where did I make the mistake?

Answer 1

Below is the fixed code:

#!/bin/ksh safeRunCommand() {   typeset cmnd="$*"   typeset ret_code    echo cmnd=$cmnd   eval $cmnd   ret_code=$?   if [ $ret_code != 0 ]; then     printf "Error: [%d] when executing command: '$cmnd'" $ret_code     exit $ret_code   fi }  command="ls -l | grep p" safeRunCommand "$command" 

Now if you look into this code, the few things that I changed are:

• use of typeset is not necessary, but it is a good practice. It makes cmnd and ret_code local to safeRunCommand
• use of ret_code is not necessary, but it is a good practice to store the return code in some variable (and store it ASAP), so that you can use it later like I did in printf "Error: [%d] when executing command: '$command'" $ret_code
• pass the command with quotes surrounding the command like safeRunCommand "$command". If you don’t then cmnd will get only the value ls and not ls -l. And it is even more important if your command contains pipes.
• you can use typeset cmnd="$*" instead of typeset cmnd="$1" if you want to keep the spaces. You can try with both depending upon how complex is your command argument.
• 'eval' is used to evaluate so that a command containing pipes can work fine

Note: Do remember some commands give 1 as the return code even though there isn't any error like grep. If grep found something it will return 0, else 1.

I had tested with KornShell and Bash. And it worked fine. Let me know if you face issues running this.

Answer 2

Try

safeRunCommand() {    "$@"     if [ $? != 0 ]; then       printf "Error when executing command: '$1'"       exit $ERROR_CODE    fi }