I have a list of directories
/home
/dir1
/dir2
...
/dir100
Some of them have no files in it. How can I use Unix find to do it?
I tried
find . -name "*" -type d -size 0
Doesn't seem to work.
Does your find have predicate -empty?
You should be able to use find . -type d -empty
If you're a zsh user, you can always do this. If you're not, maybe this will convince you:
echo **/*(/^F)
**/* will expand to every child node of the present working directory and the () is a glob qualifier. / restricts matches to directories, and F restricts matches to non-empty ones. Negating it with ^ gives us all empty directories. See the zshexpn man page for more details.
-empty reports empty leaf dirs.
If you want to find empty trees then have a look at:
http://code.google.com/p/fslint/source/browse/trunk/fslint/finded
Note that script can't be used without the other support scripts,
but you might want to install fslint and use it directly?
You can also use:
find . -type d -links 2
. and .. both count as a link, as do files.
The answer of Pimin Konstantin Kefalou prints folders with only 2 links and other files (d, f, ...).
The easiest way I have found is:
for directory in $(find . -type d); do
if [ -n "$(find $directory -maxdepth 1 -type f)" ]; then echo "$directory"
fi
done
If you have name with spaces use quotes in "$directory".
You can replace . by your reference folder.
I haven't been able to do it with one find instruction.
Related
I want to find all files with specific name "stdout.1.0", move it two/three levels up from its location. While moving it two/three levels up, I also want to rename it to "testjob.out".
All "stdout.1.0" files are located six levels down from parent directory.
./dirXXXXXX/dirXXXXXX/dirXXXXXX/dirXXXXXX/dirXXXXXX/dirXXXXXX/stdout.1.0
I used:
find . -type f -name stdout.1.0
and it outputs:
./dir100000/dir110000/dir111000/dir111100/dir111110/dir111111/stdout.1.0
./dir100000/dir110000/dir112000/dir111100/dir111110/dir111111/stdout.1.0
./dir100000/dir110000/dir113000/dir111100/dir111110/dir111111/stdout.1.0
./dir200000/dir210000/dir211000/dir211100/dir211110/dir211111/stdout.1.0
./dir200000/dir210000/dir212000/dir211100/dir211110/dir211111/stdout.1.0
./dir200000/dir210000/dir213000/dir211100/dir211110/dir211111/stdout.1.0
./dir300000/dir310000/dir311000/dir311100/dir311110/dir311111/stdout.1.0
./dir300000/dir310000/dir312000/dir311100/dir311110/dir311111/stdout.1.0
./dir300000/dir310000/dir313000/dir311100/dir311110/dir311111/stdout.1.0
.
.
./dirXXX000/dirXXX000/dirXXX000/dirXXX100/dirXXX110/dirXXX111/stdout.1.0
The directories above is just representative of where the file is, but there are multiple "stdout.1.0" files starting three levels down from parent directory.
Here is a method in plain bash using globstar shell option , without using the find:
#!/bin/bash
shopt -s globstar
for file in **/stdout.1.0; do
echo mv "$file" "${file%/*/*/*}/testjob.out"
done
Drop the echo if output looks fine.
You already know how to find them:
find . -type f -name stdout.1.0
Now, you need to move them to a higher directory (..) and rename them:
find . -type f -name stdout.1.0 -execdir mv {} ../../testjob.out \;
I would advise you to copy them first and remove later (use cp instead of mv): if anything goes wrong, you can get back easily to the current situation.
so I have the code right now down below, and I'm running into a few problems with it
I'm having trouble excluding the directories being outputted by
find ${1-.}
It is giving me the directories too instead of only names; I've tried different methods such as -prune etc.
I'm having trouble with deleting the empty files
The data given to me by
EMPTY_FILE=$(find ${1-.} -size 0)
Does not give me the correct path
Here is the output for that
TestFolder/TestFile
in this case I can't just do:
rm TestFolder/TestFile
As it is invalid path; since it needs ./TestFolder/TestFile
How would I add on the ./ or is there away to get the full path.
#!/bin/bash
echo "Here are all the files in the directory specified\n"
find ${1-.}
EMPTY_FILE=$(find ${1-.} -size 0)
echo "Here are the list of empty files\n"
echo "$EMPTY_FILE \n"
echo "Do you want to delete those empty files?(yes/no)"
read text
if [ "$text" == "yes" ]; then $(rm -- $EMPTY_FILE); fi
Any help is appreciated!
You want this:
#!/bin/bash
echo -e "Here are all the files in the directory specified\n"
# Use -printf "%f\n" to print the filename without leading directories
# Use -type f to restrict find to files
find "${1-.}" -type f -printf " %f\n"
echo -e "Here are the list of empty files\n"
# Again, use -printf "%f\n"
find "${1-.}" -type f -size 0 -printf " %f\n"
echo -e "Do you want to delete those empty files?(yes/no)"
read answer
# Delete files using the `-delete` option
[ "$answer" = "yes" ] && find "${1-.}" -type f -size 0 -delete
Also note that I've quotes "${1-.}" at all occurrences. Since it is user input, you can't rely on the input. Even if it is a path, it might still contain problematic characters, like spaces.
I'm having trouble excluding the directories being outputted by
find ${1-.}
It is giving me the directories too instead of only names
You are looking for the -type test. To instruct find to report only regular files, you could say
find ${1-.} -type f
That's probably what you really want, but what you actually asked (to exclude only directories) would be
find ${1-.} -not -type d
Excluding only directories will list symbolic links and special files, too.
in this case I can't just do:
rm TestFolder/TestFile
As it is invalid path; since it needs ./TestFolder/TestFile
Nonsense. ./TestFolder/TestFile means exactly the same thing as TestFolder/TestFile.
In any event, find does print paths starting at the specified starting path(s).
I have a feeling that I'm missing something from your question, but if all you need to do is exclude directories, just tell find to only look for files:
find . -type f -size 0 -delete
And then adjust that to suit your script. Hope this helps.
-size 0 -type f
rm with no option will not delete directories . Your claim that rm needs ./ is wrong anyway.
Here is the code that i have soo far :
echo $(pwd > adress)
var=$(head -1 adress)
rm adress
found=0 #Flag
fileshow()
{
cd $1
for i in *
do
if [ -d $i ]
then
continue
elif [ -w $i ]
then
echo $i
found=1
fi
done
cd ..
}
fileshow $1
if [ $found -eq 0 ]
then
clear
echo "$(tput setaf 1)There arent any executable files !!!$(tput sgr0)"
fi
Its working but it find files only in current directory.
I was told that i need to use some kind of recursive method to loop through all sub-directories but i dont know how to do it.
So if any one can help me i will be very grateful.
Thanks!
The effect of your script is to find the files below the current working directory that are not directories and are writeable to the current user. This can be achieved with the command:
find ./ -type f -writable
The advantage of using -type f is that it also excludes symbolic links and other special kinds of file, if that's what you want. If you want all files that are not directories (as suggested by your script), then you can use:
find ./ ! -type d -writable
If you want to sort these files (added question, assuming lexicographic ascending order), you can use sort:
find ./ -type f -writable | sort
If you want to use these sorted filenames for something else, the canonical pattern would be (to handle filenames with embedded newlines and other seldom-used characters):
while read -r -d $'\0'; do
echo "File '$REPLY' is an ordinary file and is writable"
done < <(find ./ -type f -writable -print0 | sort -z)
If you're using a very old version of find that does not support the handy -writable predicate (added to v.4.3 in 2005), then you only have file permissions to go on. You then have to be clear about what you mean by “writable” in the specific context (writable to whom?), and you can replace the -writable predicate with the -perm predicates described in #gregb's answer. If you decide that you mean “writable by anyone” you could use -perm /u=w,g=w,o=w or -perm /222, but there's actually no way of getting all the benefits of -writable just using permissions. Also note that the + form of permission tests to -perm is deprecated and should no longer be used; the / form should be used instead.
You could use find:
find /path/to/directory/ -type f -perm -o=w
Where the -o=w implies that each file has the "other write-permission" set.
or,
find /path/to/directory/ -type f -perm /u+w,g+w,o+w
Where /u+w,g+w,o+w implies that each file either has user, group, or other write-permissions set.
I need to backup all the directory hierarchy of our servers, thus I need to list all the sub directories of some of the directories in the server.
The problem is that one of those sub directories contains tens of thousands of sub directories (file with only the names of the sub directories could take couple of hundreds megabytes and the respective find command takes very long).
For example, if I have a directory A and one sub directory A/a that contains tens of thousands of sub directories, I want to use the find command to list all the sub directories of A excluding all the sub directories of A/a but not excluding A/a itself.
I tried many variations of -prune using the answers in this question to no avail.
Is there a way to use the find command in UNIX to do this?
UPDATE:
the answer by #devnull worked very well, but now i have another problem, so i will refine my question a little:
i used the following command:
find /var/www -type d \( ! -wholename "/var/www/web-release-data/*" ! -wholename "/var/www/web-development-data/*" \)
the new problem that arises is that find for some reason is still traversing the whole directory tree of "/var/www/web-release-data/" and "/var/www/web-development-data/", thus it's very slow, and I fear it could take hours.
Is there any way make find completely exclude those directories and not traverse their respective directory hierarchies?
The following should work for you:
find A -type d \( ! -wholename "A/a/*" \)
This would list all subdirectories of A including A/a but excluding subdirectories of A/a.
Example:
$ mkdir -p A/{a..c}/{1..4}
$ find A -type d \( ! -wholename "A/a/*" \)
A
A/c
A/c/4
A/c/2
A/c/3
A/c/1
A/a
A/b
A/b/4
A/b/2
A/b/3
A/b/1
Another solution:
find A \! -path "A/a/*"
If you don't want a as well, use
find A \! -path "A/a/*" -a \! -path "A/a"
Have you tried rsync(1)? It has an option --exclude=PATTERN which might work well here:
rsync -avz --exclude=A/a <source> <target>
Using rsync you wouldn't need to use find(1)
To exclude 2 subdirs:
find . -type d ! -wholename "dir/name/*" -a ! -wholename "dir/name*"
To answer your updated question, you can do
find /var/www -wholename "/var/www/web-release-data/*" -o -wholename "/var/www/web-development-data/*" -prune -o -type d -print
How can I remove all .txt files present in several directories
Dir1 >
Dir11/123.txt
Dir12/456.txt
Dir13/test.txt
Dir14/manifest.txt
In my example I want to run the remove command from Dir1.
I know the linux command rm, but i don't know how can I make this works to my case.
PS.: I'm using ubuntu.
To do what you want recursively, find is the most used tool in this case. Combined with the -delete switch, you can do it with a single command (no need to use -exec (and forks) in find like other answers in this thread) :
find Dir1 -type f -name "*.txt" -delete
if you use bash4, you can do too :
( shopt -s globstar; rm Dir1/**/*.txt )
We're not going to enter sub directories so no need to use find; everything is at the same level. I think this is what you're looking for: rm */*.txt
Before you run this you can try echo */*.txt to see if the correct files are going to be removed.
Using find would be useful if you want to search subfolders of subfolders, etc.
There is no Dir1 in the current folder so don't do find Dir1 .... If you run the find from the prompt above this will work:
find . -type f -name "*.txt" -delete