Set a command to a variable in bash script problem

You can only execute exactly one command stored in a variable. The pipe is passed as an argument to the first grep.

Example

$ printArgs() { printf %s\\n "$@"; }
# Two commands. The 1st command has parameters "a" and "b".
# The 2nd command prints stdin from the first command.
$ printArgs a b | cat
a
b
$ cmd='printArgs a b | cat'
# Only one command with parameters "a", "b", "|", and "cat".
$ $cmd
a
b
|
cat

How to do this better?

Don't execute the command using variables. Use a function.

$ cmd() { grep -i nosuid /etc/fstab | grep -iq nfs; }
$ cmd
$ echo $?
1

Solution to the actual problem

I see three options to your actual problem:

• Use a DEBUG trap and the BASH_COMMAND variable inside the trap.
• Enable bash's history feature for your script and use the hist command.
• Use a function which takes a command string and executes it using eval.

Regarding your comment on the last approach: You only need one function. Something like

execAndLog() {
    description="$1"
    shift
    if eval "$*"; then
        info="PASSED: $description: $*"
        passed+=("${FUNCNAME[1]}")
    else
       info="FAILED: $description: $*"
       failed+=("${FUNCNAME[1]}")
    done
}

You can use this function as follows

execAndLog 'Scanned system' 'grep -i nfs /etc/fstab | grep -iq noexec'

The first argument is the description for the log, the remaining arguments are the command to be executed.