222

I want to test if a directory doesn't contain any files. If so, I will skip some processing.

I tried the following:

if [ ./* == "./*" ]; then
    echo "No new file"
    exit 1
fi

That gives the following error:

line 1: [: too many arguments

Is there a solution/alternative?

1

26 Answers 26

295
if [ -z "$(ls -A /path/to/dir)" ]; then
   echo "Empty"
else
   echo "Not Empty"
fi

Also, it would be cool to check if the directory exists before.

ls -A means list all but . or ..

12
  • 12
    I had trouble getting this method to work when the /path/to/dir contained spaces and needed to be quoted. I used [ $(ls -A "$path" | wc -l) -ne 0], inspired by @ztank1013's answer.
    – pix
    May 19, 2015 at 3:23
  • 2
    For those who are looking for a one liner : [ "$(ls -A ./path/to/dir)" ] && echo 'NOT EMPTY' || echo 'EMPTY'
    – tdhulster
    Jun 3, 2016 at 16:29
  • 3
    Just in case if someone will be looking for "correct/stable" one liner: [ -z "$(ls -A /path/to/dir)" ] && { echo "Not Empty" ; YourCommandA ; true ; } || { echo "Empty" ; YourCommandB ; }. Oct 12, 2017 at 23:24
  • 5
    This checks whether the directory exists, and deals with spaces in the path (notice the nested quotes in the subshell): if [ -d "/path/to/dir" ] && [ -n "$(ls -A "/path/to/dir")" ]; then echo "Non-empty folder" else echo "Empty or not a folder" fi
    – Jonathan H
    Jan 10, 2019 at 0:40
  • 2
    Honestly, at this point, if it's my own system I just write a quick 10 line Python script, symlink it to /usr/local/bin or something, and then call it like if isNotEmpty "$directory"; then ... fi
    – bjd2385
    Aug 27, 2019 at 19:02
42

No need for counting anything or shell globs. You can also use read in combination with find. If find's output is empty, you'll return false:

if find /some/dir -mindepth 1 -maxdepth 1 | read; then
   echo "dir not empty"
else
   echo "dir empty"
fi

This should be portable.

4
  • Nice solution, but I think your echo calls reflect the wrong result : in my test (under Cygwin) find . -mindepth 1 | read had a 141 error code in a non-empty dir, and 0 in an empty dir Dec 20, 2017 at 9:17
  • @LucasCimon Not here (macOS and GNU/Linux). For an non-empty directory, read returns 0, and for an empty one, 1.
    – slhck
    Dec 20, 2017 at 11:28
  • 10
    PSA does not work with set -o pipefail Sep 10, 2019 at 2:07
  • For my case empty or not exist is the same. Can add 2>/dev/null before the pipe to avoid showing error when folder does not exist No such file or directory.
    – Martin P.
    Mar 14 at 16:55
36
if [ -n "$(find "$DIR_TO_CHECK" -maxdepth 0 -type d -empty 2>/dev/null)" ]; then
    echo "Empty directory"
else
    echo "Not empty or NOT a directory"
fi
7
  • 5
    It needs quotes (2x) and the test -n to be correct and safe (test with directory with spaces in the name, test it with non-empty directory with name '0 = 1'). ... [ -n "$(find "$DIR_TO_CHECK" -maxdepth 0 -type d -empty 2>/dev/null)" ]; ...
    – Zrin
    Mar 7, 2017 at 23:51
  • 2
    @ivan_pozdeev That's not true, at least for GNU find. You may be thinking of grep. serverfault.com/questions/225798/… Jun 19, 2018 at 10:55
  • 4
    It might be simpler to write find "$DIR_TO_CHECK" -maxdepth 0 -type d -empty | grep ., and rely on the exit status from grep. Whichever way you do it, this is very much the right answer to this question. Jul 6, 2018 at 13:56
  • 2
    I like that this doesn't require listing/reading/traversing all files in the directory. Unlike "ls" based solutions this should work much faster when the directory has lots of files. Oct 3, 2020 at 18:24
  • 2
    Found this question from a search, discarded poor answers until i found this one, tried it, found it didn't work with GNU find, went to add a comment to that effect ... and found i'd already done that two years ago. Although it should be grep -q .. Thanks again, Mr Udvari! Oct 14, 2020 at 20:11
22
#!/bin/bash
if [ -d /path/to/dir ]; then
    # the directory exists
    [ "$(ls -A /path/to/dir)" ] && echo "Not Empty" || echo "Empty"
else
    # You could check here if /path/to/dir is a file with [ -f /path/to/dir]
fi
2
  • 6
    That must be it, no need for parsing ls output, just see if it is empty or not. Using find just feels like an overkill to me. Sep 18, 2013 at 12:34
  • man test reveals that test "x" is equivalent to test -n "x"
    – aleb
    Oct 28, 2023 at 15:52
21

With FIND(1) (under Linux and FreeBSD) you can look non-recursively at a directory entry via "-maxdepth 0" and test if it is empty with "-empty". Applied to the question this gives:

if test -n "$(find ./ -maxdepth 0 -empty)" ; then
    echo "No new file"
    exit 1
fi
4
  • 2
    It may not be 100% portable, but it's elegant. Nov 28, 2018 at 2:59
  • 2
    This also finishes early in large directories, works with pipefail: set -o pipefail; { find "$DIR" -mindepth 1 || true ; } | head -n1 | read && echo NOTEMPTY || echo EMPTY
    – macieksk
    Nov 24, 2019 at 12:29
  • Was hoping to find a flag that made find return non-zero when not finding results, but this is probably the closest we're getting to that. Would at least be nice with an equivalent to ! -empty -exit 1 for GNU though. Jun 1, 2021 at 10:44
  • 1
    Finally a correct answer. No attempts at parsing ls output, not doing anything extra (like listing all files, then counting them), etc. Excellent answer. Dec 6, 2021 at 16:17
13

What about testing if directory exists and not empty in one if statement

if [[ -d path/to/dir && -n "$(ls -A path/to/dir)" ]]; then 
  echo "directory exists and is not empty"
else
  echo "directory doesn't exist or is empty"
fi
1
  • 1
    note the echo statements don't match the logic of your test (completely) Jun 14, 2022 at 14:49
10

Use the following:

count="$( find /path -mindepth 1 -maxdepth 1 | wc -l )"
if [ $count -eq 0 ] ; then
   echo "No new file"
   exit 1
fi

This way, you're independent of the output format of ls. -mindepth skips the directory itself, -maxdepth prevents recursively defending into subdirectories to speed things up.

1
  • Of course, you're now dependent on wc -l and find output format (which is reasonably plain though).
    – Daniel Beck
    Oct 31, 2011 at 10:25
9

A hacky, but bash-only, PID-free way:

is_empty() {
    test -e "$1/"* 2>/dev/null
    case $? in
        1)   return 0 ;;
        *)   return 1 ;;
    esac
}

This takes advantage of the fact that test builtin exits with 2 if given more than one argument after -e: First, "$1"/* glob is expanded by bash. This results in one argument per file. So

  • If there are no files, the asterisk in test -e "$1"* does not expand, so Shell falls back to trying file named *, which returns 1.

  • ...except if there actually is one file named exactly *, then the asterisk expands to well, asterisk, which ends up as the same call as above, ie. test -e "dir/*", just this time returns 0. (Thanks @TrueY for pointing this out.)

  • If there is one file, test -e "dir/file" is run, which returns 0.

  • But if there are more files than 1, test -e "dir/file1" "dir/file2" is run, which bash reports it as usage error, i.e. 2.

case wraps the whole logic around so that only the first case, with 1 exit status is reported as success.

Possible problems I haven't checked:

  • There are more files than number of allowed arguments--I guess this could behave similar to case with 2+ files.

  • Or there is actually file with an empty name--I'm not sure it's possible on any sane OS/FS.

7
  • 2
    Minor correction: if there is no file in dir/, then test -e dir/* is called. If the only file is '*' in dir then test will return 0. If there are more files, then it returns 2. So it works as described.
    – TrueY
    Sep 7, 2018 at 11:43
  • You're right, @TrueY, I've incorporated it in the answer. Thanks! Sep 11, 2018 at 14:45
  • 2
    This won't work if the directory contains a single file named *. A pathological case, but I've seen worse. Aug 28, 2021 at 3:02
  • @kkm it did work for me: hastebin.com/hevumumoma.txt -- Bash version 5.1.0(1)-release Sep 10, 2021 at 11:44
  • OTOH, if it does not work on older Bash (let me know) one could address that case by adding test -e "$1/*" && return 1 as the first line. Sep 10, 2021 at 11:47
7

This will do the job in the current working directory (.):

[ `ls -1A . | wc -l` -eq 0 ] && echo "Current dir is empty." || echo "Current dir has files (or hidden files) in it."

or the same command split on three lines just to be more readable:

[ `ls -1A . | wc -l` -eq 0 ] && \
echo "Current dir is empty." || \
echo "Current dir has files (or hidden files) in it."

Just replace ls -1A . | wc -l with ls -1A <target-directory> | wc -l if you need to run it on a different target folder.

Edit: I replaced -1a with -1A (see @Daniel comment)

5
  • 3
    use ls -A instead. Some file systems don't have . and .. symbolic links according to the docs.
    – Daniel Beck
    Oct 31, 2011 at 10:12
  • 2
    Thanks @Daniel, I edited my answer after your suggestion. I know the "1" might be removed too.
    – ztank1013
    Oct 31, 2011 at 10:21
  • 4
    It doesn't hurt, but it's implied if output it not to a terminal. Since you pipe it to another program, it's redundant.
    – Daniel Beck
    Oct 31, 2011 at 10:24
  • 2
    -1 is definitely redundant. Even if ls will not print one item per line when it will be piped then it doesn't affect the idea of checking if it produced zero or more lines. Oct 13, 2017 at 9:31
  • Try that on an NFS mount in a directory with 1-2 million files. You'll be pleasantly surprised that a simple empty-or-not check takes 30 min. Sticking head into the pipeline, like (( $( ls -1A | head -n1 | wc -l ) == 0 )) may be a fix, if only partial. Aug 28, 2021 at 3:09
5

Using an array:

files=( * .* )
if (( ${#files[@]} == 2 )); then
    # contents of files array is (. ..)
    echo dir is empty
fi
4
  • 6
    Very nice solution, but note that it requires shopt -s nullglob
    – xebeche
    Jan 16, 2017 at 11:24
  • 3
    The ${#files[@]} == 2 assumption doesn't stand for the root dir (you will probably not test if it's empty but some code that doesn't know about that limitation might). Jan 21, 2018 at 4:29
  • 2
    @ivan_pozdeev: What do you mean? When I do cd / && files=(* .*), I get an enumeration of all the files and directories in the root directory, which includes . and ... So the ${#files[@]} == 2 test is valid. Feb 3, 2020 at 4:08
  • Yes, this is ambiguous when you don't know whether nullglob is set (returns 3 array entries in an empty dir when unset). And as long as you're setting nullglob, you may as well set dotglob as well, as noted in the BashFAQ for this issue, then you can just use files=(*) and test against 0. Jun 14, 2022 at 20:00
2

I think the best solution is:

files=$(shopt -s nullglob; shopt -s dotglob; echo /MYPATH/*)
[[ "$files" ]] || echo "dir empty" 

thanks to https://stackoverflow.com/a/91558/520567

This is an anonymous edit of my answer that might or might not be helpful to somebody: A slight alteration gives the number of files:

files=$(shopt -s nullglob dotglob; s=(MYPATH/*); echo ${s[*]}) 
echo "MYPATH contains $files files"

This will work correctly even if filenames contains spaces.

2
  • Using an array instead of invoking a subshell will make this even better. This answer does that. Mar 6, 2020 at 21:17
  • 2
    @codeforester, the solution needs to know what are current shell opts. So one would need code to set opts and revert if needed (or figure out opts and count according to them). This will bloat the code and make it less readable. If this part of the code is on a performance critical path, then it might be worth it (or not, needs testing). For a normal use case where you check one directory, I think this answer is safe, short and self contained. If you don't need portability between scripts (which possibly use different shell opts), then the other answer is good. Mar 8, 2020 at 10:10
2

Another find solution, (which doesn't rely on other tools), and should be pretty fast:

if (( "$(find /directory/to/check/ -mindepth 1 -printf '1\n' -quit)" )); then
  echo The directory is not empty
else
  echo The directory is empty
fi
2

There are lots of good answers here for simple cases, but this QA ranks very highly in a web search, and many of the answers have subtle failures that may be important for some use cases:

  • permissions problems (i.e. reporting empty dir when it's only unsearchable)
  • ambiguity about symlinks (test dereferences, find doesn't by default, ls may depend on whether the path ends with /)
  • answers using shell globbing may rely on options like nullglob that need to be tested or explicitly set

Here is a function that is efficient for a directory with lots of files, doesn't rely on shell options for globbing, and explicitly tests for permission problems:

function mtdir {
    # test for empty directory
    if [[ -d "$1" && -r "$1" && -x "$1" ]]; then

        if find -L -- "$1" -maxdepth 0 -type d -empty | grep -q .; then
            # empty directory
            return 0
        else
            # non-empty directory
            return 1
        fi

    else
        # not a directory, not readable, or not searchable
        # if this is unexpected, might want to print error and exit
        return 2
    fi
}

This function dereferences (follows) valid symlinks, testing the targets rather than the links themselves (both at the -d test, and using find -L). If symbolic links should be tested as link files rather than their target for your application, an explicit test should be added before -d (e.g. ! -L "$1" && ...), and find -P should be used instead of find -L.

The function can be used with:

if mtdir some/dir; then
    echo "directory is empty"
    # further processing
fi

This has been tested and works under Linux and macOS, with edge cases of non-directory files, symlinks, and directories with locked down permissions (i.e. chmod 0000 testdir).

You can also test for a non-empty directory explicitly, or an error, using the return code:

mtdir some/dir
ec=$?
if [[ $ec -eq 1 ]]; then
    echo "non-empty directory"
    # further processing
elif [[ $ec -eq 2 ]]; then
    echo "unexpected error: directory not readable or not searchable"
    # further processing
fi
1
if find "${DIR}" -prune ! -empty -exit 1; then
    echo Empty
else
    echo Not Empty
fi

EDIT: I think that this solution works fine with gnu find, after a quick look at the implementation. But this may not work with, for example, netbsd's find. Indeed, that one uses stat(2)'s st_size field. The manual describes it as:

st_size            The size of the file in bytes.  The meaning of the size
                   reported for a directory is file system dependent.
                   Some file systems (e.g. FFS) return the total size used
                   for the directory metadata, possibly including free
                   slots; others (notably ZFS) return the number of
                   entries in the directory.  Some may also return other
                   things or always report zero.

A better solution, also simpler, is:

if find "${DIR}" -mindepth 1 -exit 1; then
    echo Empty
else
    echo Not Empty
fi

Also, the -prune in the 1st solution is useless.

EDIT: no -exit for gnu find.. the solution above is good for NetBSD's find. For GNU find, this should work:

if [ -z "`find \"${DIR}\" -mindepth 1 -exec echo notempty \; -quit`" ]; then
    echo Empty
else
    echo Not Empty
fi
1
1

This solution is using only shell built-ins:

function is_empty() {
  typeset dir="${1:?Directory required as argument}"
  set -- ${dir}/*
  [ "${1}" == "${dir}/*" ];
}

is_empty /tmp/emmpty && echo "empty" || echo "not empty"
1
[ $(ls -A "$path" 2> /dev/null | wc -l) -eq 0 ] && echo "Is empty or not exists." || echo "Not is empty."
1
  • While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value. Jun 19, 2020 at 17:44
1

Without calling utils like ls, find, etc.:

Inside dir:

[ "$(echo *)x" != '*x' ] || [ "$(echo .[^.]*)x" != ".[^.]*x" ] || echo "empty dir"

The idea:

  • echo * lists non-dot files
  • echo .[^.]* lists dot files except of "." and ".."
  • if echo finds no matches, it returns the search expression, i.e. here * or .[^.]* - which both are no real strings and and have to be concatenated with e.g. a letter to coerce a string
  • || alternates the possibilities in a short circuit: there is at least one non-dot file or dir OR at least one dot file or dir OR the directory is empty - on execution level: "if first possibility fails, try next one, if this fails, try next one"; here technically Bash "tries to execute" echo "empty dir", put your action for empty dirs here (eg. exit).

Checked with symlinks, yet to check with more exotic possible file types.

2
  • this is a nice one, basically uses the globs suggested in the BashFAQ here -- perhaps wrap it in a function to make it more digestable? Also not sure whether the shoptions listed in the FAQ are needed Jun 14, 2022 at 17:56
  • This test misses files named with 2 leading dots, for example ..a
    – db-inf
    Apr 15, 2023 at 11:08
1

I could be mistaken, but I think this check should be sufficient:

[[ -s /path/to/dir ]] && echo "Dir not empty" || echo "Dir empty"
1
  • 1
    -1: Bash 5.0 and 5.1.4 (Debian) return true for rm -r test 2>/dev/null; mkdir test; [[ -s test ]] && echo "true"
    – xebeche
    Jul 26, 2022 at 12:49
0

This is all great stuff - just made it into a script so I can check for empty directories below the current one. The below should be put into a file called 'findempty', placed in the path somewhere so bash can find it and then chmod 755 to run. Can easily be amended to your specific needs I guess.

#!/bin/bash
if [ "$#" == "0" ]; then 
find . -maxdepth 1 -type d -exec findempty "{}"  \;
exit
fi

COUNT=`ls -1A "$*" | wc -l`
if [ "$COUNT" == "0" ]; then 
echo "$* : $COUNT"
fi
0

For any directory other than the current one, you can check if it's empty by trying to rmdir it, because rmdir is guaranteed to fail for non-empty directories. If rmdir succeeds, and you actually wanted the empty directory to survive the test, just mkdir it again.

Don't use this hack if there are other processes that might become discombobulated by a directory they know about briefly ceasing to exist.

If rmdir won't work for you, and you might be testing directories that could potentially contain large numbers of files, any solution relying on shell globbing could get slow and/or run into command line length limits. Probably better to use find in that case. Fastest find solution I can think of goes like

is_empty() {
    test -z $(find "$1" -mindepth 1 -printf X -quit)
}

This works for the GNU and BSD versions of find but not for the Solaris one, which is missing every single one of those find operators. Love your work, Oracle.

1
  • Not a good idea. The OP simply wanted to test if the directory was empty or not. May 2, 2018 at 14:05
0

This work for me, to check & process files in directory ../IN, considering script is in ../Script directory:

FileTotalCount=0

    for file in ../IN/*; do
    FileTotalCount=`expr $FileTotalCount + 1`
done

if test "$file" = "../IN/*"
then

    echo "EXITING: NO files available for processing in ../IN directory. "
    exit

else

  echo "Starting Process: Found ""$FileTotalCount"" files in ../IN directory for processing."

# Rest of the Code
0

I made this approach:

CHECKEMPTYFOLDER=$(test -z "$(ls -A /path/to/dir)"; echo $?)
if [ $CHECKEMPTYFOLDER -eq 0 ]
then
  echo "Empty"
elif [ $CHECKEMPTYFOLDER -eq 1 ]
then
  echo "Not Empty"
else
  echo "Error"
fi
0

The Question was:

if [ ./* == "./*" ]; then
    echo "No new file"
    exit 1
fi

Answer is:

if ls -1qA . | grep -q .
    then ! exit 1
    else : # Dir is empty
fi
0

I might have missed an equivalent to this, which works on Unix

cd directory-concerned
ls * > /dev/null 2> /dev/null

return-code (test value of $?) will be 2 if nothing or 0 something found.

Note this ignores any '.' files and will probably return 2 if any of these exist without any other 'normal' filenames.

0

More solutions with find

# Tests that a directory is empty.
# Will print error message if not empty to stderr and set return
# val to non-zero (i.e. evaluates as false)
#
function is_empty() {
    find $1 -mindepth 1   -exec false {} + -fprintf /dev/stderr "%H is not empty\n" -quit
    # prints error when dir is not empty to stderr
    # -fprintf /dev/stderr "%H is not empty\n"
    #
    # -exec false {} +
    # sets the return value (i.e. $?) to indicate error
    #
    # --quit
    # terminate after the first match

}

examples

#!/bin/bash
set -eE # stop execution upon error

function is_empty() {
    find $1 -mindepth 1   -exec false {} + -fprintf /dev/stderr "%H is not empty\n" -quit
}


trap 'echo FAILED' ERR
#trap "echo DONE" EXIT

# create a sandbox to play in
d=$(mktemp -d)
f=$d/blah # this will be a potention file

set -v # turn on debugging

# dir should be empty
is_empty $d

# create a file in the dir
touch $f
! is_empty $d

# this will cause the script to fail because the dir is not empty
is_empty $d

# this line will not execute
echo "we should not get here"

output

[root@sysresccd ~/sandbox]# ./test

# dir should be empty
is_empty $d

# create a file in the dir
touch $f
! is_empty $d
/tmp/tmp.aORTHb3Trv is not empty

# this will cause the script to fail because the dir is not empty
is_empty $d
/tmp/tmp.aORTHb3Trv is not empty
echo FAILED
FAILED
1
0

Although there are many reasonable solutions here, I am personally not a big fan of most of these answers, as many return a lot of output when returning directory contents. I was expecting a solution that would better handle directories with large numbers of files, while also one that I think is easy to understand.

So, this is what I ended up with, and thought I would share:

This appears to work OK for me on RedHat:

dir="/tmp/my_empty_dir"
[[ -d "${dir}" && -z "$(find "${dir}" -not -path "${dir}" -print -quit)" ]] && echo "${dir} is empty"

In this example:

First ensure dir exists -d "$dir", otherwise this will return empty (we will also see an error sent to stderr).

However it's likely you would need to test for this separately, as a "not empty" result is likely to mean "contains files" (which is not correct)

AND

Find: (find $dir -not -path $dir -print -quit):

  • Find everything in $dir
  • Exclude the directory $dir from the resulting output
  • Print the first result (something else within $dir)
  • Quit immediately (only return the first result).

BEWARE the -path parameter takes a "pattern", so if you are expecting special characters (eg: *, [, ]) these would need to be escaped Eg:

dir='/tmp/test[dir]'
dirpath='/tmp/test\[dir\]'
find "${dir}" -not -path "${dirpath}" -print -quit

During my test, this also successfully found hidden files. ($dir/.hidden)

Find returns 0 regardless of whether anything is found, and I don't currently see a simpler way to test this, so:

As per other examples I also wrapped this in:

Empty: [[ -z "$result" ]] to test if the result is blank.

NOT Empty: [[ ! -z "$result" ]] to test if the result is not blank.

Yes, the braces around ${dir} are not really required, but I thought it best to help handle this use case

dir="/tmp/"
[[ -d "${dir}subdir" ...

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