#!/bin/bash

DIR=/var/www/tee

for i in $DIR/*
do
if [ -e $i ] && [ $(echo $?) -eq "0 " ]

then
echo The Directory is NOT EMPTY

else

echo The directory is EMPTY

fi
done

Declare the variable ,before you use it.

src=”/Users/xxxx/scripts/dirtest”
for dir in `ls -R “$src”`
do
if [ -d "$dir" ]; then
echo this is a dir: $dir
fi
done

thanks

best way is to do as below:
dir_full_path=”/tmp/foo”
if [ -d "$dir_full_path" ] ; then
if [ 4 -ge $(du -s "$dir_full_path" ] ; then
echo “empty directory $dir_full_path”
else
echo “non-empty directory $dir_full_path”
fi
fi

This snippet assumes that a directory size, by its own name, is always 4096 bytes, approximated to 4.0K evaluated to int as 4

if [ "$(ls -A $DIR)" ]; then

with

if [ "$(ls -A "$DIR")" ]; then

That’s just another pair of quotes, no escapes. Should work.

find . -type d -empty -exec touch {}/.empty \;

#!/bin/bash

DIR=”home/dan/not empty”
if [ "$(ls -A $DIR)" ]; then
echo “$DIR is not empty”
else
echo “$DIR is empty”
fi

DIR=/home/user
[ $(( $(ls -a1 $DIR | wc -l) >= 3 )) = 0 ] && echo Empty || echo Nope

That’s a one and not an ‘l’ to the ls command. An empty directory only has the two entries – . & ..

FILES=”`echo .* *`”
if [ $FILES = '. .. *' ] ; then
: empty dir
else
: not empty
fi

It is not working for me. I’m using Debian + Bash 3. It returns 3 when directory is empty, it should be 2 as empty directory has only . and ..

Any idea?

set - `echo .* *`
if [ $# = "2" ] ; then
: empty directory
else
: not empty ...
fi

David
UNIX Administratosphere