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 working the way I want. Where I made 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 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.