pass a command as an argument to bash script

Question

How do I pass a command as an argument to a bash script? In the following script, I attempted to do that, but it's not working!

#! /bin/sh

if [ $# -ne 2 ]
then
    echo "Usage: $0 <dir> <command to execute>"
    exit 1;
fi;

while read line
do
    $($2) $line
done < $(ls $1);

echo "All Done"

A sample usage of this script would be

./myscript thisDir echo

Executing the call above ought to echo the name of all files in the thisDir directory.

Answer 1

your command "echo" command is "hidden" inside a sub-shell from its argments in $line.

I think I understand what your attempting in with $($2), but its probably overkill, unless this isn't the whole story, so

 while read line ; do
    $2 $line
 done < $(ls $1)

should work for your example with thisDir echo. If you really need the cmd-substitution and the subshell, then put you arguments so they can see each other:

   $($2 $line)

And as D.S. mentions, you might need eval before either of these.

IHTH