Get exit code for command in bash/ksh

Below is the fixed code:

#!/bin/kshsafeRunCommand() {  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 few things that I changed are:

• use of typeset is not necessary but a good practice. It make cmnd and ret_code local to safeRunCommand
• use of ret_code is not necessary but a good practice to store 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 dont 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 command containing pipes can work fine

NOTE: Do remember some commands give 1 as return code even though there is no error like grep. If grep found something it will return 0 else 1.

I had tested with KSH/BASH. And it worked fine. Let me know if u face issues running this.

Try

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

It should be $cmd instead of $($cmd). Works fine with that on my box.

Edit: Your script works only for one-word commands, like ls. It will not work for "ls cpp". For this to work, replace cmd="$1"; $cmd with "$@". And, do not run your script as command="some cmd"; safeRun command, run it as safeRun some cmd.

Also, when you have to debug your bash scripts, execute with '-x' flag. [bash -x s.sh].