138

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?

2

23 Answers 23

189
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.

p.s.

ls -A means list all but . or ..

12
  • 14
    Don't use && and || simultaneously! If echo "Not Empty" fails, echo "Empty" will run! Try echo "test" && false || echo "fail"! Yes, I know echo will not fail but if you change any other command, you will be suprise! – uzsolt Oct 31 '11 at 9:27
  • 5
    Please, provide at least one example when the code above won't work. Concretely this code is absolutely correct. I hope that asker is able to adapt this for his own purposes – Andrey Atapin Oct 31 '11 at 9:56
  • 4
    where's touch /empty in my line? – Andrey Atapin Oct 31 '11 at 10:01
  • 4
    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 '15 at 3:23
  • 10
    Oh no! There is a very important problem with this code. It should be if [ -n "$(ls -A /path/to/dir)" ]; then ... Please update the answer before someone pastes this code into their server somewhere and a hacker figures out how to exploit it. If /path/to/dir isn't empty, then the filenames there get passed as arguments to /bin/test which is clearly not intended. Just add the -n argument, problem solved. Thanks! – Edward Ned Harvey Dec 5 '16 at 21:20
32

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.

8
  • 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 – Lucas Cimon Dec 20 '17 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 '17 at 11:28
  • 6
    PSA does not work with set -o pipefail – Colonel Thirty Two Sep 10 '19 at 2:07
  • using -maxdepth 1 will help with performance, too (test on /... no need to go that deep). Otherwise, why use | read instead of test -n $( find )? Not that it's bad, but I think it's the 1st time I see read used this way. – user3459474 Aug 6 '20 at 13:37
  • The options should be a -mindepth 1 -maxdepth 1 instead, because you don't want to print the first dir, but at the same time you don't want to recurse further into it. – Hi-Angel Apr 21 at 9:53
25
if [ -n "$(find "$DIR_TO_CHECK" -maxdepth 0 -type d -empty 2>/dev/null)" ]; then
    echo "Empty directory"
else
    echo "Not empty or NOT a directory"
fi
6
  • Correct and fast. Nice! – l0b0 Feb 24 '12 at 11:00
  • 4
    It needs quotes (2x) and the test -n to be correct and safe (test with directory with spaces in the name, test it with non-empty directory with name '0 = 1'). ... [ -n "$(find "$DIR_TO_CHECK" -maxdepth 0 -type d -empty 2>/dev/null)" ]; ... – Zrin Mar 7 '17 at 23:51
  • 1
    @ivan_pozdeev That's not true, at least for GNU find. You may be thinking of grep. serverfault.com/questions/225798/… – Vladimir Panteleev Jun 19 '18 at 10:55
  • It might be simpler to write find "$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. – Tom Anderson Jul 6 '18 at 13:56
  • 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. – oᴉɹǝɥɔ Oct 3 '20 at 18:24
17
#!/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
1
  • 4
    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. – akostadinov Sep 18 '13 at 12:34
11

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 "$(find ./ -maxdepth 0 -empty)" ; then
    echo "No new file"
    exit 1
fi
3
  • 1
    It may not be 100% portable, but it's elegant. – Craig Ringer Nov 28 '18 at 2:59
  • This also finishes early in large directories, works with pipefail: set -o pipefail; { find "$DIR" -mindepth 1 || true ; } | head -n1 | read && echo NOTEMPTY || echo EMPTY – macieksk Nov 24 '19 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. – Steen Schütt Jun 1 at 10:44
6

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 '11 at 10:25
5

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)

4
  • 2
    use ls -A instead. Some file systems don't have . and .. symbolic links according to the docs. – Daniel Beck Oct 31 '11 at 10:12
  • 1
    Thanks @Daniel, I edited my answer after your suggestion. I know the "1" might be removed too. – ztank1013 Oct 31 '11 at 10:21
  • 3
    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 '11 at 10:24
  • 1
    -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. – Victor Yarema Oct 13 '17 at 9:31
5

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"
else
  echo "directory doesn't exist"
fi
5

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.

2
  • 1
    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 '18 at 11:43
  • You're right, @TrueY, I've incorporated it in the answer. Thanks! – Alois Mahdal Sep 11 '18 at 14:45
4

Using an array:

files=( * .* )
if (( ${#files[@]} == 2 )); then
    # contents of files array is (. ..)
    echo dir is empty
fi
3
  • 5
    Very nice solution, but note that it requires shopt -s nullglob – xebeche Jan 16 '17 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). – ivan_pozdeev Jan 21 '18 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. – Scott Feb 3 '20 at 4:08
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. – codeforester Mar 6 '20 at 21:17
  • 1
    @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. – akostadinov Mar 8 '20 at 10:10
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"
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. – roaima May 2 '18 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
[ $(ls -A "$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. – Donald Duck Jun 19 '20 at 17:44
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
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" ...
-2

You can try to remove the directory and wait to an error; rmdir will not delete the directory if it is not empty.

_path="some/path"
if rmdir $_path >/dev/null 2>&1; then
   mkdir $_path        # create it again
   echo "Empty"
else
   echo "Not empty or doesn't exist"
fi
3
  • 4
    -1 This is the kind of code that backfires. rmdir will fail if I have no permission to remove the directory; or if it's a Btrfs subvolume; or if it belongs to a read-only filesystem. And if rmdir doesn't fail and mkdir runs: what if the already removed directory belonged to another user? what about its (possibly non-standard) permissions? ACL? extended attributes? All lost. – Kamil Maciorowski Apr 8 '18 at 21:37
  • 1
    Well, I'm just learning bash and I thought that it could be a faster way rather than iterate through all the directory, but CPUs are powerful and you are right, not secure. – impxd Apr 9 '18 at 5:21
  • -1. Destroying something to check its status is never a great idea! – codeforester Mar 6 '20 at 21:18

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