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發表者： 冷日 發表時間： 2016/8/16 2:32:33
Bash scripting: test for empty directory
I want to test if a directory doesn't contain any files. If so, I will skip some processing.
I tried the following:
That gives the following error:
Is there a solution/alternative?
Also, it would be cool to check if the directory exists before.
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:
This should be portable.
This will do the job in the current working directory (.):
or the same command split on three lines just to be more readable:
Just replace ls -1A . | wc -l with ls -1A | wc -l if you need to run it on a different target folder.
Edit: I replaced -1a with -1A (see @Daniel comment)
Use the following:
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.
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.
I think the best solution is:
thanks to http://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:
This will work correctly even if filenames contains spaces.
Using an array:
A hacky, but bash-only, PID-free way:
This takes advantage of the fact that test builtin exits with 2 if given more than one argument after -e: First, "$1"/* glob is expanded by bash. This results in one argument per file. So
if there are no files, test -e "" is run, which obviously always fails, i.e. returns 1 for "no".
If there is one file, test -e "dir/file" is run, which returns 0.
But if there are more files than 1, bash reports it as usage error, i.e. 2.
case wraps the whole logic around so that only the first case, with 1 exit status is reported as success.
Possible problems I haven't checked:
There are more files than number of allowed arguments--I guess this could behave similar to case with 2+ files.
Or there is actually file with an empty name--I'm not sure it's possible on any sane OS/FS.
原文出處：Bash scripting: test for empty directory - Super User*/
I want to test if a directory doesn't contain any files. If so, I will skip some processing.
I tried the following:
if [ ./* == "./*" ]; then
echo "No new file"
exit 1
fiThat gives the following error:
line 1: [: too many arguments
Is there a solution/alternative?
if [ "$(ls -A /path/to/dir)" ]; then
echo "Not Empty"
else
echo "Empty"
fi
Also, it would be cool to check if the directory exists before.
if [ $(find $DIR_TO_CHECK -maxdepth 0 -type d -empty 2>/dev/null) ]; then
echo "Empty directory"
else
echo "Not empty or NOT a directory"
fiNo need for counting anything or shell globs. You can also use read in combination with find. If find's output is empty, you'll return false:
if find /some/dir -mindepth 1 | read; then
echo "dir not empty"
else
echo "dir empty"
fi
This should be portable.
This will do the job in the current working directory (.):
[ `ls -1A . | wc -l` -eq 0 ] && echo "Current dir is empty." || echo "Current dir has files (or hidden files) in it."
or the same command split on three lines just to be more readable:
[ `ls -1A . | wc -l` -eq 0 ] && \
echo "Current dir is empty." || \
echo "Current dir has files (or hidden files) in it."
Just replace ls -1A . | wc -l with ls -1A
Edit: I replaced -1a with -1A (see @Daniel comment)
#!/bin/bash
if [ -d /path/to/dir ]; then
# the directory exists
[ "$(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]
fiUse the following:
count="$( find /path -mindepth 1 -maxdepth 1 | wc -l )"
if [ $count -eq 0 ] ; then
echo "No new file"
exit 1
fi
This way, you're independent of the output format of ls. -mindepth skips the directory itself, -maxdepth prevents recursively defending into subdirectories to speed things up.
This is all great stuff - just made it into a script so I can check for empty directories below the current one. The below should be put into a file called 'findempty', placed in the path somewhere so bash can find it and then chmod 755 to run. Can easily be amended to your specific needs I guess.
#!/bin/bash
if [ "$#" == "0" ]; then
find . -maxdepth 1 -type d -exec findempty "{}" \;
exit
fi
COUNT=`ls -1A "$*" | wc -l`
if [ "$COUNT" == "0" ]; then
echo "$* : $COUNT"
fiI think the best solution is:
files=$(shopt -s nullglob; shopt -s dotglob; echo /MYPATH/*)
[[ "$files" ]] || echo "dir empty"
thanks to http://stackoverflow.com/a/91558/520567
This is an anonymous edit of my answer that might or might not be helpful to somebody: A slight alteration gives the number of files:
files=$(shopt -s nullglob dotglob; s=(MYPATH/*); echo ${s[*]})
echo "MYPATH contains $files files"This will work correctly even if filenames contains spaces.
Using an array:
files=( * .* )
if (( ${#files[@]} == 2 )); then
# contents of files array is (. ..)
echo dir is empty
fiA 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, test -e "" is run, which obviously always fails, i.e. returns 1 for "no".
If there is one file, test -e "dir/file" is run, which returns 0.
But if there are more files than 1, 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.
原文出處：Bash scripting: test for empty directory - Super User*/