I have recently run into a problem.
I used a utility to move all my music files into directories based on tags. This left a LOT of almost empty folders. The folders, in general, contain a thumbs.db file or some sort of image for album art. The mp3s have the correct album art in their new directories, so the old ones are okay to delete.
Basically, I need to find any directories within D:/Music/ that:
-Do not have any subdirectories
-Do not contain any mp3 files
And then delete them.
I figured this would be easier to do in a shell script or bash script or whatever else linux/unix world than in Windows 8.1 (HAHA).
Any suggestions? I'm not very experienced writing scripts like this.
This should get you started
find /music -mindepth 1 -type d |
while read dt
do
find "$dt" -mindepth 1 -type d | read && continue
find "$dt" -iname '*.mp3' -type f | read && continue
echo DELETE $dt
done
Here's the short story...
find . -name '*.mp3' -o -type d -printf '%h\n' | sort | uniq > non-empty-dirs.tmp
find . -type d -print | sort | uniq > all-dirs.tmp
comm -23 all-dirs.tmp non-empty-dirs.tmp > dirs-to-be-deleted.tmp
less dirs-to-be-deleted.tmp
cat dirs-to-be-deleted.tmp | xargs rm -rf
Note that you might have to run all the commands a few times (depending on your repository's directory depth) before you're done deleting all recursive empty directories...
And the long story goes...
You can approach this problem from two basic perspective: either you find all directories, then iterate over each of them, check if it contain any mp3 file or any subdirectory, if not, mark that directory for deletion. It will works, but on large very large repositories, you might expect a significant run time.
Another approach, which is in my sense much more interesting, is to build a list of directories NOT to be deleted, and subtract that list from the list of all directories. Let's work the second strategy, one step at a time...
First of all, to find the path of all directories that contains mp3 files, you can simply do:
find . -name '*.mp3' -printf '%h\n' | sort | uniq
This means "find any file ending with .mp3, then print the path to it's parent directory".
Now, I could certainly name at least ten different approaches to find directories that contains at least one subdirectory, but keeping the same strategy as above, we can easily get...
find . -type d -printf '%h\n' | sort | uniq
What this means is: "Find any directory, then print the path to it's parent."
Both of these queries can be combined in a single invocation, producing a single list containing the paths of all directories NOT to be deleted.. Let's redirect that list to a temporary file.
find . -name '*.mp3' -o -type d -printf '%h\n' | sort | uniq > non-empty-dirs.tmp
Let's similarly produce a file containing the paths of all directories, no matter if they are empty or not.
find . -type d -print | sort | uniq > all-dirs.tmp
So there, we have, on one side, the complete list of all directories, and on the other, the list of directories not to be deleted. What now? There are tons of strategies, but here's a very simple one:
comm -23 all-dirs.tmp non-empty-dirs.tmp > dirs-to-be-deleted.tmp
Once you have that, well, review it, and if you are satisfied, then pipe it through xargs to rm to actually delete the directories.
cat dirs-to-be-deleted.tmp | xargs rm -rf
Related
I have found the following which will list the files in all subdirectories, hide the last 5, and then delete the rest:
find -type f -printf '%T# %P\n' | sort -n | cut -d' ' -f2- | head -n -5 | xargs rm
Unfortunately if I don't know how many subdirectories there are, it won't delete the correct number of files. Does anyone have a way to transverse each directory, and then delete all but the newest of file in each subdirectory?
Directory structure would be the following:
-> Base Directory -> Parent Directory -> Child directory
I'd write a script.
It would be a recursive function:
call function: rm_files(base_dir)
list all directories
if there are directories go through on the list and call rm_files(act_dir) for each item
else (if there is no directories):
list all files
delete all files but the newest
return from function
In case lot of subdirectories it may be memory problem because of the recursive function.
I found I was able to do what I needed to do with the following one liner:
find . -name *.* -mmin +59 -delete > /dev/null
I'm looking for a specific directory file count that returns a number. I would type it into the terminal and it can give me the specified directory's file count.
I've already tried echo find "'directory' | wc -l" but that didn't work, any ideas?
You seem to have the right idea. I'd use -type f to find only files:
$ find some_directory -type f | wc -l
If you only want files directly under this directory and not to search recursively through subdirectories, you could add the -maxdepth flag:
$ find some_directory -maxdepth 1 -type f | wc -l
Open the terminal and switch to the location of the directory.
Type in:
find . -type f | wc -l
This searches inside the current directory (that's what the . stands for) for all files, and counts them.
The fastest way to obtain the number of files within a directory is by obtaining the value of that directory's kMDItemFSNodeCount metadata attribute.
mdls -name kMDItemFSNodeCount directory_name -raw|xargs
The above command has a major advantage over find . -type f | wc -l in that it returns the count almost instantly, even for directories which contain millions of files.
Please note that the command obtains the number of files, not just regular files.
I don't understand why folks are using 'find' because for me it's a lot easier to just pipe in 'ls' like so:
ls *.png | wc -l
to find the number of png images in the current directory.
I'm using tree, this is the way :
tree ph
I am writing a shell script where it is checking if the bin directory is present under all the users directory under /home directory. The bin directory can be present directly under user directory or under the child directory of the user directory.
I mean let say I have a user as amit under /home. So the bin directory can be present directly as /amit/bin or can be present as /amit/jash/bin
Now my requirement is that I should have a list of users directories where the bin directory is not present either directly under user directory or under the child directory of the user directory. I tried the command as :
find /home -type d ! -exec test -e '{}/bin' \; -print
but it is not working. However when I am replacing the bin directory with some file, the command is working fine. Looks like this command is particularly for files. Is there any similar command for directories?? Any help on this will be greatly appreciated.
You're on the right track. The catch is that your test of "does the following directory NOT exist in this target" can't be expressed within find's conditions in such a way as to return only the top-level directory. So you need to nest, one way or another.
One strategy would be to use a for loop in bash:
$ mkdir foo bar baz one two
$ mkdir bar/bin baz/bin
$ for d in /home/*/; do find "$d" -type d -name bin | grep -q . || echo "$d"; done
foo/
one/
two/
This uses pathname expansion (globbing) to generate the list of directories to test, and then checks for the existence of "bin". If that check fails (i.e. find outputs nothing), the directory is printed. Note the trailing slash on /home/*/, which ensures that you will only be searching within directories, rather than files that might accidentally exist in /home/.
Another possibility might be to use nested finds, if you don't want to depend on bash:
$ find /home/ -type d -depth 1 -not -exec sh -c "find {}/ -type d -name bin -print | grep -q . " \; -print
/home/foo
/home/one
/home/two
This roughly duplicates the effect of the bash for loop above, but by nesting find within find -exec. It uses grep -q . to convert the output of find into an exit status that can be used as a condition for the outer find.
Note that since you're looking for a bin directory, we want to use test -d rather than test -e (which would also check for a bin file, which probably does not matter to you.)
Another option is to use bash process redirection. On multiple lines for easier reading:
cd /home/
comm -3 \
<(printf '%s\n' */ | sed 's|/.*||' | sort) \
<(find */ -type d -name bin | cut -d/ -f1 | uniq)
This unfortunately requires you to change to the /home directory before running, because of the way it strips off subdirectories. You can of course collapse this into a big long one-liner if you feel so inclined.
This comm solution also has the risk of failing on directories with special characters in their names, like newlines.
One last option is bash-only but more than a one-liner. It involves subtracting the directories containing "bin" from the full list. It uses an associative array and globstar, so it depends on bash version 4.
#!/usr/bin/env bash
shopt -s globstar
# Go to our root
cd /home
# Declare an associative array
declare -A dirs=()
# Populate the array with our "full" list of home directories
for d in */; do dirs[${d%/}]=""; done
# Remove directories that contain a "bin" somewhere inside 'em
for d in **/bin; do unset dirs[${d%%/*}]; done
# Print the result in reproducible form
declare -p dirs
# Or print the result just as a list of words.
printf '%s\n' "${!dirs[#]}"
Note that we're storing directories in the array index, which (1) makes it easy for us to find and delete items, and (2) insures unique entries, even if one user has multiple "bin" directories under their home.
cd /home
find . -maxdepth 1 -type d ! -name . | sort > a
find . -type d -name bin | cut -d/ -f1,2 | sort > b
comm -23 a b
Here, I'm making two sorted lists. The first contains all the home directories, and the second contains the top parent of any bin subdirectory. Finally I output any items from the first list not present in the second.
I have 2 different recursive directories, in one directory have 200 .txt files in another have 210 .txt files, need a script to find the different file names and remove them from the directory.
There are probably better ways, but I think about:
find directory1 directory2 -name \*.txt -printf '%f\n' |
sort | uniq -u |
xargs -I{} find directory1 directory2 -name {} -delete
find directory1 directory2 -name \*.txt -printf '%f\n':
print basename of each file matching the glob *.txt
sort | uniq -u:
only print unique lines (if you wanted to delete duplicate, it would have been uniq -d)
xargs -I{} find directory1 directory2 -name {} -delete:
remove them (re-specify the path to narrow the search and avoid deleting files outside the initial search path)
Notes
Thank's to #KlausPrinoth for all the suggestions.
Obviously I'm assuming a GNU userland, I suppose people running with the tools providing bare minimum POSIX compatibility will be able to adapt it.
Yet another way is to use diff which is more than capable in finding file differences in files in directories. For instance if you have d1 and d2 that contain your 200 and 210 files respectively (with the first 200 files being the same), you could use diff and process substitution to provide the names to remove to a while loop:
( while read -r line; do printf "rm %s\n" ${line##*: }; done < <(diff -q d1 d2) )
Output (of d1 with 10 files, d2 with 12 files)
rm file11.txt
rm file12.txt
diff will not fit all circumstances, but is does a great job finding directory differences and is quite flexible.
I need to write a recursive script to delete all folders in a subfolder named 'date-2012-01-01_12_30' but leave the two latest.
/var/www/temp/updates/ then hundreds of folders by 'date' and by 'code'
e.g.
/var/www/temp/updates/2012-01-01/temp1/date-2012-01-_12_30
/var/www/temp/updates/2012-01-01/temp1/date-2012-02-_13_30
/var/www/temp/updates/2012-01-01/temp1/date-2013-11-_12_30
/var/www/temp/updates/2012-01-01/temp2/date-2012-01-_12_30
I was thinking about using a find to get the folder but unsure how to know what folders I can delete as the script will have to know how date- folders are in that subfolder and which ones are the latest ones
Hmm, any help would be great?
This should work:
find /var/www/temp/updates/ -type d -name "date-*" -printf '%T# %p\n' | sort -n | head -n -2 | cut -f2- | xargs rm -rf
find prints out the directory paths along with their last modification times. This is then sorted and all but the last two are deleted.
If all folders are in subdirectories temp1, temp2, ..., you can just use ls -tr
ls -dtr /var/www/temp/updates/2012-01-01/temp*/* | head -n -2 | xargs rm -rf
This lists all folders sorted by time ls -dtr, takes all but the two latest head and removes the remaining folders xargs rm -rf.