bash variable substitution within command substitution

Question

I want to do something like the following:

#!/bin/bash

cmd="find . -name '*.sh'"
echo $($cmd)

What I expect is that it will show all the shell script files in the current directory, but nothing happened.

I know that I can solve the problem with eval according to this post

#!/bin/bash

cmd="find . -name '*.sh'"
eval $cmd

So my question is why command substitution doesn't work here and what's the difference between $(...) and eval in terms of the question?

Answer 1

Command substitution works here. Just you have wrong quoting. Your script find only one file name! This one with single quotes and asteriks in it:

'*.sh'

You can create such not usual file by this command and test it:

touch "'*.sh'"

Quoting in bash is different than in other programming languages. Check out details in this answer.

What you need is this quoting:

cmd="find . -name *.sh"
echo $($cmd)

Answer 2

Since you are already including the patter *.sh inside double quotes, there's no need for the single quotes to protect the pattern, and as a result the single quotes are part of the pattern.

You can try using an array to keep *.sh quoted until it is passed to the command substitution:

cmd=(find . -name '*.sh')
echo $("${cmd[@]}")

I don't know if there is a simple way to convert your original string to an array without the pattern being expanded.

Update: This isn't too bad, but it's probably better to just create the array directly if you can.

cmd="find . -name *.sh"
set -f
cmd=($cmd)
set +f
echo $("${cmd[@]}")