Checking from shell script if a directory contains

Question

From a shell script, how do I check if a directory contains files?

Something similar to this

if [ -e /some/dir/* ]; then echo "huzzah"; fi;

but which works if the directory contains one or several files (the above one only works with exactly 0 or 1 files).

Answer 1

Mixing prune things and last answers, I got to

find "$some_dir" -prune -empty -type d | read && echo empty || echo "not empty"

that works for paths with spaces too

Answer 2

Try:

if [ ! -z `ls /some/dir/*` ]; then echo "huzzah"; fi

Answer 3

if ls /some/dir/* >/dev/null 2>&1 ; then echo "huzzah"; fi;

Answer 4

I am surprised the wooledge guide on empty directories hasn't been mentioned. This guide, and all of wooledge really, is a must read for shell type questions.

Of note from that page:

Never try to parse ls output. Even ls -A solutions can break (e.g. on HP-UX, if you are root, ls -A does the exact opposite of what it does if you're not root -- and no, I can't make up something that incredibly stupid).

In fact, one may wish to avoid the direct question altogether. Usually people want to know whether a directory is empty because they want to do something involving the files therein, etc. Look to the larger question. For example, one of these find-based examples may be an appropriate solution:

   # Bourne
   find "$somedir" -type f -exec echo Found unexpected file {} \;
   find "$somedir" -maxdepth 0 -empty -exec echo {} is empty. \;  # GNU/BSD
   find "$somedir" -type d -empty -exec cp /my/configfile {} \;   # GNU/BSD

Most commonly, all that's really needed is something like this:

   # Bourne
   for f in ./*.mpg; do
        test -f "$f" || continue
        mympgviewer "$f"
    done

In other words, the person asking the question may have thought an explicit empty-directory test was needed to avoid an error message like mympgviewer: ./*.mpg: No such file or directory when in fact no such test is required.

Answer 5

I dislike the ls - A solutions posted. Most likely you wish to test if the directory is empty because you don't wish to delete it. The following does that. If however you just wish to log an empty file, surely deleting and recreating it is quicker then listing possibly infinite files?

This should work...

if !  rmdir ${target}
then
    echo "not empty"
else
    echo "empty"
    mkdir ${target}
fi

Answer 6

dir_is_empty() {
   [ "${1##*/}" = "*" ]
}

if dir_is_empty /some/dir/* ; then
   echo "huzzah"
fi

Assume you don't have a file named * into /any/dir/you/check, it should work on bash dash posh busybox sh and zsh but (for zsh) require unsetopt nomatch.

Performances should be comparable to any ls which use *(glob), I guess will be slow on directories with many nodes (my /usr/bin with 3000+ files went not that slow), will use at least memory enough to allocate all dirs/filenames (and more) as they are all passed (resolved) to the function as arguments, some shell probably have limits on number of arguments and/or length of arguments.

A portable fast O(1) zero resources way to check if a directory is empty would be nice to have.

update

The version above doesn't account for hidden files/dirs, in case some more test is required, like the is_empty from Rich’s sh (POSIX shell) tricks:

is_empty () (
cd "$1"
set -- .[!.]* ; test -f "$1" && return 1
set -- ..?* ; test -f "$1" && return 1
set -- * ; test -f "$1" && return 1
return 0 )

But, instead, I'm thinking about something like this:

dir_is_empty() {
    [ "$(find "$1" -name "?*" | dd bs=$((${#1}+3)) count=1 2>/dev/null)" = "$1" ]
}

Some concern about trailing slashes differences from the argument and the find output when the dir is empty, and trailing newlines (but this should be easy to handle), sadly on my busybox sh show what is probably a bug on the find -> dd pipe with the output truncated randomically (if I used cat the output is always the same, seems to be dd with the argument count).