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How can I list the current directory or any directory path contents without using ls command? Can we do it using echo command?

marked as duplicate by jimmij, chaos, Stephen Kitt, Drav Sloan, Archemar Aug 11 '15 at 12:29

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  • found it! " echo * " will do the job. – kashminder Aug 11 '15 at 11:33
printf '%s\n' *

as a shell command will list the non-hidden files in the current directory, one per line. If there's no non-hidden file, it will display * alone except in those shells where that issue has been fixed (csh, tcsh, fish, zsh, bash -O failglob).

echo *

Will list the non-hidden files separated by space characters except (depending on the shell/echo implementation) when the first file name starts with - or file names contain backslash characters.

It's important to note that it's the shell expanding that * into the list of files before passing it to the command. You can use any command here like, head -- * to display the first few lines (with those head implementations that accept several files), stat -- *...

I you want to include hidden files:

printf '%s\n' .* *

(depending on the shell, that will also include . and ..). With zsh:

printf '%s\n' *(D)

Among the other applications (beside shell globs and ls) that can list the content of a directory, there's also find:

find . ! -name . -prune

(includes hidden files except . and ..).

On Linux, lsattr (lists the Linux extended file attributes):

lsattr -a # to include hidden files like with ls

If you just want a list of directory contents: find . -maxdepth 1

or for any other dir: find <dir> -maxdepth 1

  • 1
    Note that it includes hidden files except ... -maxdepth is a GNU extension (now supported by a few other implementations); the portable/standard equivalent would be: find . -name . -o -prune or find . ! -name . -prune if you don't care for the . entry). – Stéphane Chazelas Aug 11 '15 at 12:41

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