[bash] Get exit code for command in bash/ksh

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?

There are several things wrong with your script.

Functions (subroutines) should be declared before attempting to call them. You probably want to return() but not exit() from your subroutine to allow the calling block to test the success or failure of a particular command. That aside, you don't capture 'ERROR_CODE' so that is always zero (undefined).

It's good practice to surround your variable references with curly braces, too. Your code might look like:

#!/bin/sh
command="/bin/date -u"          #...Example Only

safeRunCommand() {
   cmnd="$@"                    #...insure whitespace passed and preserved
   $cmnd
   ERROR_CODE=$?                #...so we have it for the command we want
   if [ ${ERROR_CODE} != 0 ]; then
      printf "Error when executing command: '${command}'\n"
      exit ${ERROR_CODE}        #...consider 'return()' here
   fi
}

safeRunCommand $command
command="cp"
safeRunCommand $command

The normal idea would be to run the command and then use $? to get the exit code. However, some times you have multiple cases in which you need to get the exit code. For example, you might need to hide it's output but still return the exit code, or print both the exit code and the output.

ec() { [[ "$1" == "-h" ]] && { shift && eval $* > /dev/null 2>&1; ec=$?; echo $ec; } || eval $*; ec=$?; }

This will give you the option to suppress the output of the command you want the exit code for. When the output is suppressed for the command, the exit code will directly be returned by the function.

I personally like to put this function in my .bashrc file

Below I demonstrate a few ways in which you can use this:

# In this example, the output for the command will be
# normally displayed, and the exit code will be stored
# in the variable $ec.

$ ec echo test
test
$ echo $ec
0

# In this example, the exit code is output
# and the output of the command passed
# to the `ec` function is suppressed.

$ echo "Exit Code: $(ec -h echo test)"
Exit Code: 0

# In this example, the output of the command
# passed to the `ec` function is suppressed
# and the exit code is stored in `$ec`

$ ec -h echo test
$ echo $ec
0

Solution to your code using this function

#!/bin/bash
if [[ "$(ec -h 'ls -l | grep p')" != "0" ]]; then
    echo "Error when executing command: 'grep p' [$ec]"
    exit $ec;
fi

You should also note that the exit code you will be seeing will be for the grep command that's being run, as it is the last command being executed. Not the ls.

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].