Show only hidden files (dot files) in ls alias

Question

I'm using the command

ls -a | grep '^\.'

for showing only the hidden files. I added the line

alias hidden='ls -a | grep '^\.'' # show only hidden files

to .bash_aliases file

but this does not work. It's probably the problem with ' character.

Could you please help me write the correct alias?

with examples that are even more complex, you very rarely have to switch from single- to double-quotes. This can be done just by butting the quotes up against one another. For this example, it would be `'ls -a | grep '"'"'^\.'"'"'` It looks awful (and for this example it is completely unnecessary because you can just swap out double-quotes for either of the pairs), but on very rare occasions it is useful. You might also want `-C` flag (force multi-column output). See this image http://imgur.com/a/VIVFP — Dylan, Sep 29 '16 at 00:21

score 130 · Answer 1 · answered Sep 28 '11 at 21:49

130

Have the shell list the dot files, and tell ls not to see through directories:

ls -d .*

answered Sep 28 '11 at 21:49

Gilles 'SO- stop being evil'

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This has the advantage of allowing multi-column listing, unlike the `grep`-based approach. – anol Oct 12 '14 at 13:17
1

However, shows also directories (as mentioned). Any way to keep it simple as is and show only files? – honzajde Sep 10 '16 at 09:47
2

@honzajde The question didn't ask to exclude directories. If you want to exclude directories, you can use `ls -d .* | grep -v '^d'` or (zsh only) `print -lr .*(.)` – Gilles 'SO- stop being evil' Sep 10 '16 at 21:39
this not work for me, on Ubuntu 16.04 – Andrew_1510 Oct 08 '17 at 08:41
This has the advantage of showing directories as well – young_souvlaki Aug 23 '20 at 22:07

score 33 · Accepted Answer · answered Sep 28 '11 at 18:38

Either make the inner pair of quotes double quotes:

alias hidden='ls -a | grep "^\."'

Or make the outer pair of quotes double quotes:

alias hidden="ls -a | grep '^\.'"

Or make all quotes double quotes and escape the inner pair:

alias hidden="ls -a | grep \"^\.\""

Or make it a function, so you can pass some arguments when calling:

hidden() { ls -a "$@" | grep '^\.'; }

score 12 · Answer 3 · edited Sep 17 '12 at 01:38

12

ls -Ad .* #This will list all the hidden files & directories while retaining the color & formatting

OR

To create an alias of the same:

alias lh='ls -Ad .*'

OR

Same thing could be done via grep command and pipe operator; however it would loose the color and formatting:

ls -a|grep "^\."

OR

Via alias:

alias lh='ls -a|grep "^\."'

edited Sep 17 '12 at 01:38

jasonwryan

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answered Aug 26 '12 at 10:44

Prince Bhanwra

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score 4 · Answer 4 · answered Sep 28 '11 at 18:38

4

You can use double quotes:

alias hidden="ls -a | grep '^\.'"

or concatenate more single quoted strings

alias hidden='ls -a | grep '\''^\.'\'

or remove at all internal quotes

alias hidden='ls -a | grep ^\\.'

answered Sep 28 '11 at 18:38

enzotib

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rahmu · Answer 5 · 2011-09-28T18:54:11.863

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For the record this doesn't seem to work with me, since ls -a prints two (sometimes more columns). I would recommend using the -1 option to make sure every file is in its own line. Something like this:

alias hidden='ls -a1 | grep "^\."'

edited Sep 28 '11 at 18:54

answered Sep 28 '11 at 18:44

rahmu

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I just tried `ls -a | cat` it still outputs on 2 columns. I should mention I'm using Solaris and not the GNU ls. Maybe that's why. In all cases your `-1` solution worked perfectly and is much more elegant than my hackish workaround. I'm updating my answer. – rahmu Sep 28 '11 at 18:53
I probably missed this discussion, but look here if something will go wrong. – xralf Sep 28 '11 at 19:31

score 2 · Answer 6 · answered Nov 16 '20 at 21:32

First answer which shows an alias allowing to add a directory path as usual

Try

alias lsh='ls -al --ignore="[^.]*"'

where ls ignores all files and directories which NOT start with a .

Or

alias lsh='ls -Al --ignore="[^.]*"'

to avoid . and .. entries

Then just call

lsh /path/directory/

score 1 · Answer 7 · answered May 05 '17 at 07:58

Making it slightly more complicated, but avoiding parsing ls.

llsh () { find "${@:-.}/" -maxdepth 1 -type f -name ".*" -ls; }

lsh () { find "${@:-.}/" -maxdepth 1 -type f -name ".*" -print; }

The two shell function will use find to generate a list of all hidden regular files in the current directory, or in the directory given on the command line.

The llsh function will generate a "long listing" which will be only slightly more verbose than ls -l, while lsh generates a single-column listing like ls -1.

Arcticstar · Answer 8 · 2022-12-10T06:13:54.907

Here's a function version of this instead of an alias. It does rely on extended glob, which can be turned on with shopt -s extglob. This allows for SHELL pattern matching, which in this implementation prevents . and .. from being returned as entries.

hidden() {
    if [[ $# -eq 1 ]]; then
        exec 3>&1
        _hidden=$(cd "${1}"; ls -d .!(|.)?* 1>&3)
        unset _hidden
        exec 3>&-
    elif [[ $# -gt 1 ]]; then
        exec 3>&1
        for i in $@; do
            _hidden=$(cd "${i}"; echo "${i}:" 1>&3; ls -d .!(|.)* 1>&3)
        done
        unset _hidden i
        exec 3>&-
    else
        ls -d .!(|.)*
    fi
}

It cds into the desired directory in a subshell, and redirects the output to the current shell. This stops the path of the provided directory being written preceding each entry. It should behave very similarly to ls when given directory arguments.

If you want to show . and .., just change the three ls commands in the function to ls -d .* which is also not dependant on extended glob. You can also add other arguments to this command, like --color=auto. The function itself will not accept these option arguments. If options are already included in an alias redefining ls, the function will use the alias and therefore implement those options.

score 0 · Answer 9 · answered Sep 29 '11 at 03:16

0

Does your ls support -A? From man ls:

    -a, --all
          do not ignore entries starting with .

   -A, --almost-all
          do not list implied . and ..

$ ls --version
ls (GNU coreutils) 8.5
Copyright (C) 2010 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>.
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  
Written by Richard M. Stallman and David MacKenzie.

answered Sep 29 '11 at 03:16

waltinator

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Yes, it does, but I wanted `only` hidden files. – xralf Sep 29 '11 at 06:44
`ls -A -1 -d -F .* | egrep -v '/$'` – waltinator Sep 30 '11 at 21:01

score 0 · Answer 10 · edited May 04 '17 at 22:34

0

find . -type f -name ".*"

OR

find . -type f -name ".*" | while read file; do basename $file; done

edited May 04 '17 at 22:34

Stephen Rauch

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answered May 04 '17 at 22:14

kemiko

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Hi! You could improve your answer by explaining in detail how it works, and also by mentioning why, in your opinion, it answers the question. – dhag May 05 '17 at 00:25

score 0 · Answer 11 · answered Jul 23 '19 at 11:06

# add it to ~/.bashrc

    hidden() {  # lists hidden files and directories
        find . -maxdepth 1 -name '.?*' -exec ls --color -d {} \;
        }

    hiddenfiles() { # lists hidden files
        find . -maxdepth 1 -type f -name '.?*' -exec ls --color -d {} \;
        }

    hiddendirs() { # lists directories
        find . -maxdepth 1 -type d -name '.?*' -exec ls --color -d {} \;
        }

score -3 · Answer 12 · edited Sep 21 '16 at 13:50

-3

Just simply type the below command and dont confuse with above explained stuff.

ls -A -1 -d -F .* | egrep -v '/$'

edited Sep 21 '16 at 13:50

Archemar

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answered Sep 21 '16 at 13:02

srinath

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I think the OP asked for how to write an alias properly. – countermode Sep 21 '16 at 13:24
1

That's just a copy of waltinator's comment on his/her answer. For this answer to be useful, you need to explain how and why those particular `ls` flags and the `egrep` command answer the question. – Anthony Geoghegan Sep 21 '16 at 13:46

score -3 · Answer 13 · edited May 22 '17 at 05:09

-3

ls -lart | sed -n "/ \.[A-Za-z0-9][A-Za-z0-9]*/p"

edited May 22 '17 at 05:09

Stephen Rauch

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answered May 22 '17 at 04:51

shailesh Chanderiya

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Can you please clarify your answer, one line of code is not counted as good answer – Romeo Ninov May 22 '17 at 05:49
as per requirement he need to list only hidden files. so i have list all files including hidden files by ls -lart. then i have use regex to show that files which is start with dot(.) . I have got outpout like this. – shailesh Chanderiya Aug 30 '17 at 03:12
[g906016@hklu2574881:[PTA] ~]$ ls -lart | sed -n "/ \.[A-Za-z0-9][A-Za-z0-9]*/p" -rw-r--r-- 1 g906016 g906016 658 Dec 29 2016 .zshrc -rw-r--r-- 1 g906016 g906016 171 Dec 29 2016 .kshrc -rw------- 1 g906016 g906016 27 Dec 29 2016 .k5login -rw-r--r-- 1 g906016 g906016 176 Dec 29 2016 .bash_profile -rw-r--r-- 1 g906016 g906016 18 Dec 29 2016 .bash_logout -rwxr-xr-x 1 g906016 g906016 0 Dec 29 2016 .dir_colors drwxr-x--- 3 g906016 g906016 4096 Dec 29 2016 .subversion – shailesh Chanderiya Aug 30 '17 at 03:12

Show only hidden files (dot files) in ls alias

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