We are looking for a script that will traverse in recursive mode all subfolders and list all SVN repository URLs and the path where it was found.
It will be used on /home folder of a user.
Recursively find directories, and for each of them try to get the SVN info. If it is successfull, then don't descend into the directory and print the directory name.
find -type d -exec bash -c "svn info {} > /dev/null 2> /dev/null" \; -prune -print
This will list the directories.
If you want the repository info, you can add it in the middle of the find exec command.
find -type d -exec bash -c "svn info {} 2> /dev/null | grep URL" \; -prune -print
Edit:
I found much better results by only testing for the presence of an .svn subdirectory. Then, svn info is called once at the end and grepped for Path and URL. (Plus using -0 to prevent from spaces in filenames.)
find -type d -exec test -d "{}/.svn" \; -prune -print0 | xargs -0 svn info | grep -e '\(Path\|URL\)'
Depending on what kind of limitations you have there could be different ways. The easiest way would be to do svn ls -R | grep -v "\." to grab all the sub-folders from the repository location you're at and feed that in to a for loop that adds the URI to the root of the ls to the front of each line. This will however not be adequate if you have files that do not contain a "." as they will be detected as folders. Unfortunately svn ls doesn't allow you to filter by file/folder, so if you need to deal with filenames without extensions then you'd have to do something different such as checking out the source and using find to get the folder names.
user_home=... # fill in the user's home dir
old_dir=/../
find $user_home -name .svn | rev | cut -f2- -d/ | rev | while read line ; do
echo -n "$line"$'\t'
wc -c <<<"$line"
done | sort -t$'\t' -k1,1 -k2,2n | while read dir size ; do
if [[ $dir != $old_dir* ]] ; then
old_dir=$dir
svn info $dir | grep URL
echo PATH: $dir
echo
fi
done
Just hope users do not store SVN under directories with spaces in names.
Related
Im trying to sort thousands of files and put them in their own folder based on file extension. For example, JPG files to go into a JPG folder.
How can I create a for loop to address this?
What I have attempted to far:
# This listed out all the file extensions and the count for each extension:
find . -type f | rev | cut -d. -f1 | rev | tr '[:upper:]' '[:lower:]' | sort | uniq --count | sort -rn
#This was able to find all jpegs in the media folder
find /media -iname '*.jpg'
# This worked, however the MV command does not create the folder
find /media -iname '*.jpg' -exec mv '{}' /media/genesis/Passport/consolidated/jpg/ \; #
Im guessing that the for loop would be something like but I cant seem to figure it out:
for dir in 'find /media -iname "*.jpg"'; do
mkdir $dir;
mv $dir/*;
done
To find all the files of a particular extension and move then to a destination we can use find command as follows:
find . -name "*.jpg" -exec mv {} $destination/ \;
This being the critical logic, you can just create a for loop on top of this to iterate through all the known file extensions in your media folder
## List all the known extensions in your media folder.
declare -a arr=("jpg" "png" "svg")
## Refer question 1842254 to get all the extentsions in your folder.
## find . -type f | perl -ne 'print $1 if m/\.([^.\/]+)$/' | sort -u
## now loop through the above array
for i in "${arr[#]}"
do
echo "$i"
## Create a directory based on file extension
destination="$i"
mkdir $destination
find /media -iname "*.$i" -exec mv {} $destination/ \;
done
Note that you can change your destination to be whatever you want. I just left it in the same directory.
Here is an executable shell script for your reference: https://onlinegdb.com/NEQSpojhM
I would like to find the newest sub directory in a directory and save the result to variable in bash.
Something like this:
ls -t /backups | head -1 > $BACKUPDIR
Can anyone help?
BACKUPDIR=$(ls -td /backups/*/ | head -1)
$(...) evaluates the statement in a subshell and returns the output.
There is a simple solution to this using only ls:
BACKUPDIR=$(ls -td /backups/*/ | head -1)
-t orders by time (latest first)
-d only lists items from this folder
*/ only lists directories
head -1 returns the first item
I didn't know about */ until I found Listing only directories using ls in bash: An examination.
This ia a pure Bash solution:
topdir=/backups
BACKUPDIR=
# Handle subdirectories beginning with '.', and empty $topdir
shopt -s dotglob nullglob
for file in "$topdir"/* ; do
[[ -L $file || ! -d $file ]] && continue
[[ -z $BACKUPDIR || $file -nt $BACKUPDIR ]] && BACKUPDIR=$file
done
printf 'BACKUPDIR=%q\n' "$BACKUPDIR"
It skips symlinks, including symlinks to directories, which may or may not be the right thing to do. It skips other non-directories. It handles directories whose names contain any characters, including newlines and leading dots.
Well, I think this solution is the most efficient:
path="/my/dir/structure/*"
backupdir=$(find $path -type d -prune | tail -n 1)
Explanation why this is a little better:
We do not need sub-shells (aside from the one for getting the result into the bash variable).
We do not need a useless -exec ls -d at the end of the find command, it already prints the directory listing.
We can easily alter this, e.g. to exclude certain patterns. For example, if you want the second newest directory, because backup files are first written to a tmp dir in the same path:
backupdir=$(find $path -type -d -prune -not -name "*temp_dir" | tail -n 1)
The above solution doesn't take into account things like files being written and removed from the directory resulting in the upper directory being returned instead of the newest subdirectory.
The other issue is that this solution assumes that the directory only contains other directories and not files being written.
Let's say I create a file called "test.txt" and then run this command again:
echo "test" > test.txt
ls -t /backups | head -1
test.txt
The result is test.txt showing up instead of the last modified directory.
The proposed solution "works" but only in the best case scenario.
Assuming you have a maximum of 1 directory depth, a better solution is to use:
find /backups/* -type d -prune -exec ls -d {} \; |tail -1
Just swap the "/backups/" portion for your actual path.
If you want to avoid showing an absolute path in a bash script, you could always use something like this:
LOCALPATH=/backups
DIRECTORY=$(cd $LOCALPATH; find * -type d -prune -exec ls -d {} \; |tail -1)
With GNU find you can get list of directories with modification timestamps, sort that list and output the newest:
find . -mindepth 1 -maxdepth 1 -type d -printf "%T#\t%p\0" | sort -z -n | cut -z -f2- | tail -z -n1
or newline separated
find . -mindepth 1 -maxdepth 1 -type d -printf "%T#\t%p\n" | sort -n | cut -f2- | tail -n1
With POSIX find (that does not have -printf) you may, if you have it, run stat to get file modification timestamp:
find . -mindepth 1 -maxdepth 1 -type d -exec stat -c '%Y %n' {} \; | sort -n | cut -d' ' -f2- | tail -n1
Without stat a pure shell solution may be used by replacing [[ bash extension with [ as in this answer.
Your "something like this" was almost a hit:
BACKUPDIR=$(ls -t ./backups | head -1)
Combining what you wrote with what I have learned solved my problem too. Thank you for rising this question.
Note: I run the line above from GitBash within Windows environment in file called ./something.bash.
I've got a couple of thousand images that are saved as logs that need to be deleted.
To avoid the limit of rm and to do this across multiple servers, I used the following code
Net::SSH::Multi.start(:on_error => :ignore) do |session|
# define servers in groups for more granular access
session.group :app do
session.use 'example#example', :password=> 'example'
end
# execute commands on a subset of servers
session.with(:app).exec "find /tmp/motion -maxdepth 1 -not -name 'lastsnap.jpg' -print0 | sudo xargs -0 rm"
end
An ls -l lastsnap.jpg shows that lastsnap.jpg is linked to another file, like so
30 Jun 3 08:18 lastsnap.jpg -> 81-20140603081840-snap.jpg
This other file is constantly changed due to logging scenario that i mentioned above.
Reiterating the question, how do I delete every other logged file that is NOT lastsnap.jpg and it's linked file.
Thanks for the help :)
cd /tmp/motion
ls -1 | grep -v -E '$(basename `find . -lname lastsnap.jpg`)|lastsnap.jpg' | while read n ; do rm -rvf $n ; done
EDIT as per the comment
cd /tmp/motion; rm -rvf $(ls -1 | grep -v -E "$(basename `find . -lname lastsnap.jpg`)|lastsnap.jpg")
Note: Make sure that your file names don't have spaces in it. Other wise this method will not work and needs modification in order to accommodate spaces in the file name.
I wrote a logic using find command. Check whether its useful to you.
My directory contains following files
pyramid-stone.jpg
tallest_water_slide.jpg
SAOLA.JPG
testnap.jpg
silicon_valley_talent.jpg
The_Organic_Battery_From_Japan.jpg
Out of which testnap.jpg is a link
testnap.jpg -> pyramid-stone.jpg
So i wrote a small awk script to get the link name and where its pointing to
IG1=`ls -l | grep ^l | awk '{printf $(NF-2);}'`
IG2=`ls -l | grep ^l | awk '{printf $(NF);}'`
Then i used find command to print all jpg's instead of the link
find . -type f \( -iname "*.jpg" ! -iname $IG1 ! -iname $IG2 \)
OP is
./SAOLA.JPG
./silicon_valley_talent.jpg
./tallest_water_slide.jpg
./The_Organic_Battery_From_Japan.jpg
NOTE:You have add rm to remove files after the find command
I am able to list all the directories by
find ./ -type d
I attempted to list the contents of each directory and count the number of files in each directory by using the following command
find ./ -type d | xargs ls -l | wc -l
But this summed the total number of lines returned by
find ./ -type d | xargs ls -l
Is there a way I can count the number of files in each directory?
This prints the file count per directory for the current directory level:
du -a | cut -d/ -f2 | sort | uniq -c | sort -nr
Assuming you have GNU find, let it find the directories and let bash do the rest:
find . -type d -print0 | while read -d '' -r dir; do
files=("$dir"/*)
printf "%5d files in directory %s\n" "${#files[#]}" "$dir"
done
find . -type f | cut -d/ -f2 | sort | uniq -c
find . -type f to find all items of the type file, in current folder and subfolders
cut -d/ -f2 to cut out their specific folder
sort to sort the list of foldernames
uniq -c to return the number of times each foldername has been counted
You could arrange to find all the files, remove the file names, leaving you a line containing just the directory name for each file, and then count the number of times each directory appears:
find . -type f |
sed 's%/[^/]*$%%' |
sort |
uniq -c
The only gotcha in this is if you have any file names or directory names containing a newline character, which is fairly unlikely. If you really have to worry about newlines in file names or directory names, I suggest you find them, and fix them so they don't contain newlines (and quietly persuade the guilty party of the error of their ways).
If you're interested in the count of the files in each sub-directory of the current directory, counting any files in any sub-directories along with the files in the immediate sub-directory, then I'd adapt the sed command to print only the top-level directory:
find . -type f |
sed -e 's%^\(\./[^/]*/\).*$%\1%' -e 's%^\.\/[^/]*$%./%' |
sort |
uniq -c
The first pattern captures the start of the name, the dot, the slash, the name up to the next slash and the slash, and replaces the line with just the first part, so:
./dir1/dir2/file1
is replaced by
./dir1/
The second replace captures the files directly in the current directory; they don't have a slash at the end, and those are replace by ./. The sort and count then works on just the number of names.
Here's one way to do it, but probably not the most efficient.
find -type d -print0 | xargs -0 -n1 bash -c 'echo -n "$1:"; ls -1 "$1" | wc -l' --
Gives output like this, with directory name followed by count of entries in that directory. Note that the output count will also include directory entries which may not be what you want.
./c/fa/l:0
./a:4
./a/c:0
./a/a:1
./a/a/b:0
Slightly modified version of Sebastian's answer using find instead of du (to exclude file-size-related overhead that du has to perform and that is never used):
find ./ -mindepth 2 -type f | cut -d/ -f2 | sort | uniq -c | sort -nr
-mindepth 2 parameter is used to exclude files in current directory. If you remove it, you'll see a bunch of lines like the following:
234 dir1
123 dir2
1 file1
1 file2
1 file3
...
1 fileN
(much like the du-based variant does)
If you do need to count the files in current directory as well, use this enhanced version:
{ find ./ -mindepth 2 -type f | cut -d/ -f2 | sort && find ./ -maxdepth 1 -type f | cut -d/ -f1; } | uniq -c | sort -nr
The output will be like the following:
234 dir1
123 dir2
42 .
Everyone else's solution has one drawback or another.
find -type d -readable -exec sh -c 'printf "%s " "$1"; ls -1UA "$1" | wc -l' sh {} ';'
Explanation:
-type d: we're interested in directories.
-readable: We only want them if it's possible to list the files in them. Note that find will still emit an error when it tries to search for more directories in them, but this prevents calling -exec for them.
-exec sh -c BLAH sh {} ';': for each directory, run this script fragment, with $0 set to sh and $1 set to the filename.
printf "%s " "$1": portably and minimally print the directory name, followed by only a space, not a newline.
ls -1UA: list the files, one per line, in directory order (to avoid stalling the pipe), excluding only the special directories . and ..
wc -l: count the lines
This can also be done with looping over ls instead of find
for f in */; do echo "$f -> $(ls $f | wc -l)"; done
Explanation:
for f in */; - loop over all directories
do echo "$f -> - print out each directory name
$(ls $f | wc -l) - call ls for this directory and count lines
This should return the directory name followed by the number of files in the directory.
findfiles() {
echo "$1" $(find "$1" -maxdepth 1 -type f | wc -l)
}
export -f findfiles
find ./ -type d -exec bash -c 'findfiles "$0"' {} \;
Example output:
./ 6
./foo 1
./foo/bar 2
./foo/bar/bazzz 0
./foo/bar/baz 4
./src 4
The export -f is required because the -exec argument of find does not allow executing a bash function unless you invoke bash explicitly, and you need to export the function defined in the current scope to the new shell explicitly.
My answer is a little different, due to the options of find, you can actually be much more flexible. Just try:
find . -type f -printf "%h\n" | sort | uniq -c
With the "%h" option to "-printf", find prints only the directory of the files it found. Then sort and count with "uniq -c". This prints the number of search result entries with the same directory, per directory.
Using further options on find, you can be much more flexible. For example, to get an overview how many files in which directory have been modified at a certain date, use:
find . -newermt "2022-01-01 00:00:00" -type f -printf "%TY-%Tm-%Td %h\n" | sort | uniq -c
This finds all files that have been modified since 1. January 2022, prints (with "-printf") the modification date and the directory, then sorts and counts them. In this example, each line in the result has the number of files, the date of modification (without time), and the directory.
Note that "-printf" may not be available in all versions of find I think.
I combined #glenn jackman's answer and #pcarvalho's answer(in comment list, there is something wrong with pcarvalho's answer because the extra style control function of character '`'(backtick)).
My script can accept path as an augument and sort the directory list as ls -l, also it can handles the problem of "space in file name".
#!/bin/bash
OLD_IFS="$IFS"
IFS=$'\n'
for dir in $(find $1 -maxdepth 1 -type d | sort);
do
files=("$dir"/*)
printf "%5d,%s\n" "${#files[#]}" "$dir"
done
FS="$OLD_IFS"
My first answer in stackoverflow, and I hope it can help someone ^_^
THis could be another way to browse through the directory structures and provide depth results.
find . -type d | awk '{print "echo -n \""$0" \";ls -l "$0" | grep -v total | wc -l" }' | sh
find . -type f -printf '%h\n' | sort | uniq -c
gives for example:
5 .
4 ./aln
5 ./aln/iq
4 ./bs
4 ./ft
6 ./hot
I tried with some of the others here but ended up with subfolders included in the file count when I only wanted the files. This prints ./folder/path<tab>nnn with the number of files, not including subfolders, for each subfolder in the current folder.
for d in `find . -type d -print`
do
echo -e "$d\t$(find $d -maxdepth 1 -type f -print | wc -l)"
done
This will give the overall count.
for file in */; do echo "$file -> $(ls $file | wc -l)"; done | cut -d ' ' -f 3| py --ji -l 'numpy.sum(l)'
A super fast miracle command, which recursively traverses files to count the number of images in a directory and organize the output by image extension:
find . -type f | sed -e 's/.*\.//' | sort | uniq -c | sort -n | grep -Ei '(tiff|bmp|jpeg|jpg|png|gif)$'
Credits: https://unix.stackexchange.com/a/386135/354980
I edited the script in order to exclude all node_modules directories inside the analyzed one.
This can be used to check if the project number of files is exceeding the maximum number that the file watcher can handle.
find . -type d ! -path "*node_modules*" -print0 | while read -d '' -r dir; do
files=("$dir"/*)
printf "%5d files in directory %s\n" "${#files[#]}" "$dir"
done
To check the maximum files that your system can watch:
cat /proc/sys/fs/inotify/max_user_watches
node_modules folder should be added to your IDE/editor excluded paths in slow systems, and the other files count shouldn't ideally exceed the maximum (which can be changed though).
Easy Method:
find ./|grep "Search_file.txt" |cut -d"/" -f2|sort |uniq -c
In my case I needed the count at subfolder level, so I did:
du -a | cut -d/ -f3 | sort | uniq -c | sort -nr
Easy way to recursively find files of a given type. In this case, .jpg files for all folders in current directory:
find . -name *.jpg -print | wc -l
omg why the complex commands. just use something like
find whatever_folder | wc -l
I need to calculate a summary MD5 checksum for all files of a particular type (*.py for example) placed under a directory and all sub-directories.
What is the best way to do that?
The proposed solutions are very nice, but this is not exactly what I need. I'm looking for a solution to get a single summary checksum which will uniquely identify the directory as a whole - including content of all its subdirectories.
Create a tar archive file on the fly and pipe that to md5sum:
tar c dir | md5sum
This produces a single MD5 hash value that should be unique to your file and sub-directory setup. No files are created on disk.
find /path/to/dir/ -type f -name "*.py" -exec md5sum {} + | awk '{print $1}' | sort | md5sum
The find command lists all the files that end in .py.
The MD5 hash value is computed for each .py file. AWK is used to pick off the MD5 hash values (ignoring the filenames, which may not be unique).
The MD5 hash values are sorted. The MD5 hash value of this sorted list is then returned.
I've tested this by copying a test directory:
rsync -a ~/pybin/ ~/pybin2/
I renamed some of the files in ~/pybin2.
The find...md5sum command returns the same output for both directories.
2bcf49a4d19ef9abd284311108d626f1 -
To take into account the file layout (paths), so the checksum changes if a file is renamed or moved, the command can be simplified:
find /path/to/dir/ -type f -name "*.py" -exec md5sum {} + | md5sum
On macOS with md5:
find /path/to/dir/ -type f -name "*.py" -exec md5 {} + | md5
ire_and_curses's suggestion of using tar c <dir> has some issues:
tar processes directory entries in the order which they are stored in the filesystem, and there is no way to change this order. This effectively can yield completely different results if you have the "same" directory on different places, and I know no way to fix this (tar cannot "sort" its input files in a particular order).
I usually care about whether groupid and ownerid numbers are the same, not necessarily whether the string representation of group/owner are the same. This is in line with what for example rsync -a --delete does: it synchronizes virtually everything (minus xattrs and acls), but it will sync owner and group based on their ID, not on string representation. So if you synced to a different system that doesn't necessarily have the same users/groups, you should add the --numeric-owner flag to tar
tar will include the filename of the directory you're checking itself, just something to be aware of.
As long as there is no fix for the first problem (or unless you're sure it does not affect you), I would not use this approach.
The proposed find-based solutions are also no good because they only include files, not directories, which becomes an issue if you the checksumming should keep in mind empty directories.
Finally, most suggested solutions don't sort consistently, because the collation might be different across systems.
This is the solution I came up with:
dir=<mydir>; (find "$dir" -type f -exec md5sum {} +; find "$dir" -type d) | LC_ALL=C sort | md5sum
Notes about this solution:
The LC_ALL=C is to ensure reliable sorting order across systems
This doesn't differentiate between a directory "named\nwithanewline" and two directories "named" and "withanewline", but the chance of that occurring seems very unlikely. One usually fixes this with a -print0 flag for find, but since there's other stuff going on here, I can only see solutions that would make the command more complicated than it's worth.
PS: one of my systems uses a limited busybox find which does not support -exec nor -print0 flags, and also it appends '/' to denote directories, while findutils find doesn't seem to, so for this machine I need to run:
dir=<mydir>; (find "$dir" -type f | while read f; do md5sum "$f"; done; find "$dir" -type d | sed 's#/$##') | LC_ALL=C sort | md5sum
Luckily, I have no files/directories with newlines in their names, so this is not an issue on that system.
If you only care about files and not empty directories, this works nicely:
find /path -type f | sort -u | xargs cat | md5sum
A solution which worked best for me:
find "$path" -type f -print0 | sort -z | xargs -r0 md5sum | md5sum
Reason why it worked best for me:
handles file names containing spaces
Ignores filesystem meta-data
Detects if file has been renamed
Issues with other answers:
Filesystem meta-data is not ignored for:
tar c - "$path" | md5sum
Does not handle file names containing spaces nor detects if file has been renamed:
find /path -type f | sort -u | xargs cat | md5sum
For the sake of completeness, there's md5deep(1); it's not directly applicable due to *.py filter requirement but should do fine together with find(1).
If you want one MD5 hash value spanning the whole directory, I would do something like
cat *.py | md5sum
Checksum all files, including both content and their filenames
grep -ar -e . /your/dir | md5sum | cut -c-32
Same as above, but only including *.py files
grep -ar -e . --include="*.py" /your/dir | md5sum | cut -c-32
You can also follow symlinks if you want
grep -aR -e . /your/dir | md5sum | cut -c-32
Other options you could consider using with grep
-s, --no-messages suppress error messages
-D, --devices=ACTION how to handle devices, FIFOs and sockets;
-Z, --null print 0 byte after FILE name
-U, --binary do not strip CR characters at EOL (MSDOS/Windows)
GNU find
find /path -type f -name "*.py" -exec md5sum "{}" +;
Technically you only need to run ls -lR *.py | md5sum. Unless you are worried about someone modifying the files and touching them back to their original dates and never changing the files' sizes, the output from ls should tell you if the file has changed. My unix-foo is weak so you might need some more command line parameters to get the create time and modification time to print. ls will also tell you if permissions on the files have changed (and I'm sure there are switches to turn that off if you don't care about that).
Using md5deep:
md5deep -r FOLDER | awk '{print $1}' | sort | md5sum
I want to add that if you are trying to do this for files/directories in a Git repository to track if they have changed, then this is the best approach:
git log -1 --format=format:%H --full-diff <file_or_dir_name>
And if it's not a Git directory/repository, then the answer by ire_and_curses is probably the best bet:
tar c <dir_name> | md5sum
However, please note that tar command will change the output hash if you run it in a different OS and stuff. If you want to be immune to that, this is the best approach, even though it doesn't look very elegant on first sight:
find <dir_name> -type f -print0 | sort -z | xargs -0 md5sum | md5sum | awk '{ print $1 }'
md5sum worked fine for me, but I had issues with sort and sorting file names. So instead I sorted by md5sum result. I also needed to exclude some files in order to create comparable results.
find . -type f -print0 \
| xargs -r0 md5sum \
| grep -v ".env" \
| grep -v "vendor/autoload.php" \
| grep -v "vendor/composer/" \
| sort -d \
| md5sum
I had the same problem so I came up with this script that just lists the MD5 hash values of the files in the directory and if it finds a subdirectory it runs again from there, for this to happen the script has to be able to run through the current directory or from a subdirectory if said argument is passed in $1
#!/bin/bash
if [ -z "$1" ] ; then
# loop in current dir
ls | while read line; do
ecriv=`pwd`"/"$line
if [ -f $ecriv ] ; then
md5sum "$ecriv"
elif [ -d $ecriv ] ; then
sh myScript "$line" # call this script again
fi
done
else # if a directory is specified in argument $1
ls "$1" | while read line; do
ecriv=`pwd`"/$1/"$line
if [ -f $ecriv ] ; then
md5sum "$ecriv"
elif [ -d $ecriv ] ; then
sh myScript "$line"
fi
done
fi
If you want really independence from the file system attributes and from the bit-level differences of some tar versions, you could use cpio:
cpio -i -e theDirname | md5sum
There are two more solutions:
Create:
du -csxb /path | md5sum > file
ls -alR -I dev -I run -I sys -I tmp -I proc /path | md5sum > /tmp/file
Check:
du -csxb /path | md5sum -c file
ls -alR -I dev -I run -I sys -I tmp -I proc /path | md5sum -c /tmp/file