Set a command to a variable in bash script problem - bash

Question

Trying to run a command as a variable but I am getting strange results
Expected result "1" :
grep -i nosuid /etc/fstab | grep -iq nfs
echo $?
1
Unexpected result as a variable command:
cmd="grep -i nosuid /etc/fstab | grep -iq nfs"
$cmd
echo $?
0
It seems it returns 0 as the command was correct not actual outcome. How to do this better ?

Answer 1

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.

Answer 2

using bash -x or set -x will allow you to see what bash executes:
> cmd="grep -i nosuid /etc/fstab | grep -iq nfs"
> set -x
> $cmd
+ grep -i nosuid /etc/fstab '|' grep -iq nfs
as you can see your pipe | is passed as an argument to the first grep command.