Question

I’m learning ksh, I’m trying to run a command using a subshell, but I got different results, I’m guessing the reason.

root@setPrompt[/home/za] X=$("ls -ltr")
ksh: ls -ltr:  not found.
root@setPrompt[/home/za] X=$('ls -ltr')
ksh: ls -ltr:  not found.
root@setPrompt[/home/za] X="$(ls -ltr)"
root@setPrompt[/home/za] echo $X
total 5256 -rw-

thanks

Answer 1

$() runs the enclosed command in a subshell and returns its output. Your first two examples are trying to run the command "ls -ltr". Since you’ve quoted the entire command, the shell is going to look for a command whose entire name ls -ltr, not one whose name is ls and is being passed the options -ltr. The third example runs the command ls, with the argument -ltr and X gets the output of that command. Since the $() was enclosed by double-quotes, field splitting and pathname expansion are not performed.

An example of the difference:

$ ls
bin
$ echo $(echo 'b*')
bin
$ echo "$(echo 'b*')"
b*

See also the SUS specification for command expansion.