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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?

kashminder
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2 Answers2

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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
lsattr -a # to include hidden files like with ls
Stéphane Chazelas
  • 544,893
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If you just want a list of directory contents: find . -maxdepth 1

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

csny
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    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