I should echo only names of files or directories with this construction:

ls -Al | while read string

ls -Al output :

drwxr-xr-x  12 s162103  studs         12 march 28 12:49 personal domain
drwxr-xr-x   2 s162103  studs          3 march 28 22:32 public_html
drwxr-xr-x   7 s162103  studs          8 march 28 13:59 WebApplication1

For example if I try:

ls -Al | while read string
echo "$string" | awk '{print $9}

then output only files and directories without spaces. If file or directory have spaces like "personal domain" it will be only word "personal".

I need very simple solution. Maybe there is better solution than awk.

  • 3
    Why not just ls -Al *' '*? Parsing ls's output never leads to anything good. – manatwork Mar 30 '13 at 15:01
  • 1
    If you need something that will work in any *nix try avoiding using ls -Al | while. A simple and more reliable way is for string in *; do echo "$string"; done. – forcefsck Mar 30 '13 at 15:17
  • need only with this construction..okay,not simple,any solution! – Alex Zern Mar 30 '13 at 15:31
  • 2
    Why do you need ls? Parsing ls and "work in all *nix and will not break if something happened" do not go well together. – terdon Mar 30 '13 at 15:59
  • 2
    mywiki.wooledge.org/ParsingLs – Braiam May 10 '14 at 23:29

You really should not parse the output of ls. If this is a homework assignment and you are required to, your professor does not know what they're talking about. Why don't you do something like this:

The good...

find ./  -printf "%f\n"


for n in *; do printf '%s\n' "$n"; done

...the bad...

If you really really want to use ls, you can make it a little bit more robust by doing something like this:

ls -lA | awk -F':[0-9]* ' '/:/{print $2}'

...and the ugly

If you insist on doing it the wrong, dangerous way and just have to use a while loop, do this:

ls -Al | while IFS= read -r string; do echo "$string" | 
    awk -F':[0-9]* ' '/:/{print $2}'; done

Seriously though, just don't.

  • drwx------ 2 s162103 studs 3 oct. 9 2012 .ssh drwxr-xr-x 3 s162103 studs 6 oct. 25 09:02 .subversion drwxrwxrwt 3 s162103 studs 3 nov. 15 2012 .TempoItems sometimes i've got date without time, only year and gawk return empty – Alex Zern Mar 30 '13 at 16:30
  • 1
    @AlexZern Well, that's one of the reasons you don't parse the output of ls. – terdon Mar 30 '13 at 16:33
  • 2
    @AlexZern why do you need the -l switch anyway? If all you want is the file names, just run ls -A. Using -l just makes your life harder since you now have to parse the output. – terdon Mar 30 '13 at 16:37
  • because I need additional information like access, date, owner,etc. And I wrote a huge script ,which works well, except print file and dir names with spaces. – Alex Zern Mar 30 '13 at 16:46
  • 3
    OK then. What we have here is an XY problem. ls is just not the best way of getting what you need. Why don't you post a new question explaining the information you want to collect and we can suggest ways of doing it. Alternatively, you can use your script as is, and use one of the suggestions above to print the name only. But seriously, don't parse ls, just tell us exactly what you are trying to do in a new question. – terdon Mar 30 '13 at 16:49

Is there some reason that ls -A1* won't work?


$ touch file1 file2 file\ with\ spaces
$ ls -Al
total 0
-rw-r--r-- 1 bahamat bahamat 0 Mar 30 22:31 file1
-rw-r--r-- 1 bahamat bahamat 0 Mar 30 22:31 file2
-rw-r--r-- 1 bahamat bahamat 0 Mar 30 22:31 file with spaces
$ ls -A1
file with spaces

* Note: that's a capital letter A and the number one.


I wonder why no one mentioned this simple command:

ls -a | sort
  • 2
    Brilliant! Any idea how piping to sort causes multi-column block to be unpivoted into single column? – msciwoj Sep 14 '16 at 10:49
  • @msciwoj it just takes any strings separated by whitespaces and prints them sorted separated by endlines – Ben Sep 14 '16 at 12:32
  • 5
    It works here because ls reverts to ls -1 (single columnd output) when its output doesn't go to a terminal (and disables non-printable character quoting/replacement; @Kusalananda, your ls is not POSIX compliant if it still does). ls output is already sorted. So piping it to sort will at best do nothing, and at worse mangle the output (if file names contain newline characters) or fail for filenames that are not text. ls -a | cat would be better. Or even better ls -A1 as already mentioned (and because you generally do want the replacements/quoting when displayed on a terminal) – Stéphane Chazelas Mar 23 '18 at 11:22
  • 1
    @StéphaneChazelas Noted. ls behaves differently when output is not a terminal. – Kusalananda Mar 23 '18 at 11:25

You cannot parse the output of ls, let alone ls -l because newline, just like space is as valid a character as any in a filename. Also, you'll need to consider symlinks that have an output like ... foo -> bar.

Why would you use -l anyway if you only want the file name?

Just do:

for file in *; do

If you want to include dot files (except . and ..), depending on the shell:


for file in *(ND); do


shopt -s nullglob dotglob
for file in *; do


for file in ~(N)*; do


for file in .[!.]* ..?* *; do
  [ -e "$file" ] || [ -L "$file" ] || continue
ls -Al | tr -s ' ' | cut -f9- -d' '

compress multiple spaces into single spaces with tr then you can use cut to split on the fields

  • In my case where I knew the full path just not the file name, cut worked best. Full command for getting the most recent with version-aware sorting: ls /workdir/node-* | cut -f 3 -d / | sort -V -r | head -n 1 – Samuel Neff Feb 5 '20 at 20:23

Try (bash, zsh, ksh93 syntax):

find . -type f -print0 | while IFS= read -r -d '' filename; do

This goes recursive and lists only on normal files (i.e. no dirs or symlinks/fifos/devices...).

See also:


Probably the easiest way to print all file names in directory is just:

echo *

Downside - won't print hidden files.

  • Depending on the shell, there may be shell options to set to make the pattern expand to hidden names (e.g. dotglob in the bash shell), or special patterns that does so (like *(D) in the zsh shell). This is mentioned in another answer. – Kusalananda Mar 17 '20 at 14:16

How about:

for file in * .[!.]*
  printf "%s\n" "$file"

You can use:

ls -lA | awk '{print $9}' 

It will work if you happen not to have any files with spaces in their names.... otherwise a slightly more convoluted approach may work:

ls -lA | awk '{$1=$2=$3=$4=$5=$6=$7=$8=""; print $0;}

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