i saw a few posts in forum but i cant manage to make them work for me
i have a script that runs in a folder and i want it to count the size only of the files in that folder but without the folders inside.
so if i have
file1.txt
folder1
file2.txt
it will return the size in bytes of file1+file2 without folder1
find . -maxdepth 1 -type f
gives me a list of all the files i want to count but how can i get the size of all this files?
The tool for this is xargs:
find "$dir" -maxdepth 1 -type f -print0 | xargs -0 wc -c
Note that find -print0 and xargs -0 are GNU extensions, but if you know they are available, they are well worth using in your script - you don't know what characters might be present in the filenames in the target directory.
You will need to post-process the output of wc; alternatively, use cat to give it a single input stream, like this:
find "$dir" -maxdepth 1 -type f -print0 | xargs -0 cat | wc -c
That gives you a single number you can use in following commands.
(I've assumed you meant "size" in bytes; obviously substitute wc -m if you meant characters or wc -l if you meant lines).
I don't need to do this in one line, but I've only got 1 line so far.
find . -perm -111 +type f | sort -r
What I'm trying to do is write a bash script that will display the list of all files in the current directory that are executable (z to a). I want the script to do the same for all subdirectories. What I'm having difficulty doing is displaying the name of the subdirectory before the list of executable files in that directory / subdirectory.
So, to clarify, desirable output might look like this:
program1
program2
SubDir1
program3
SubDirSubDir2
program4
SubDir2
program5
What I have right now (the above code) does this. Its not removing /path and it isn't listing the name of the new directory when directories are changed.
./exfile
./test/exfile1
./test1/program2
./test1/program
./first
Hopefully that was clear.
This will work.
I changed the permission to -100 because maybe some programs are only executable by its owner.
for d in $(find . -type d); do
echo "in $d:"
find $d -maxdepth 1 -perm -100 -type f | sed 's#.*/##'
done
This will do the trick for you.
find . -type d | sort | xargs -n1 -I{} bash -c "find {} -type f -maxdepth 1 -executable | sort -r"
The first find command lists all directories and sub directories and sort them in ascending order.
The sorted directories/sub-directories are then passed to xargs which calls bash to find the files within the directory/sub-directory and sort them in descending order.
If you prefer to also print the directory, you may run it without -type f.
You can use find on all directories and combine it with -print (to print the directory name) and -exec (to execute a find for files in that directory):
find . -type d -print -exec bash -c 'find {} -type f -depth 1 -perm +0111 | sort -r' \;
Let's break this down. First, you have the directory search:
find . -type d -print
Then the command to execute for each directory:
find {} -type f -depth 1 -perm +0111 | sort -r
The -exec switch will expand the path wherever it sees {}. Because this uses a pipe operator that is shell syntax, the whole thing is wrapped in bash -c.
You can expand on this further. If you want to strip the directory name off the files and space our your results nicer, something like this might suffice:
find {} -type f -depth 1 -print0 -perm +0111 | xargs -n1 -0 basename | sort -r && echo
Hmm, the sorting requirement makes this tricky - the "for d in $(find...)" command is clever, but hard to control the sorting. How about this? Everything is z->a, including the directories, but the awk statement is a bit of a monster ;_)
find `pwd` -perm 111 -type f |
sort -r |
xargs -n1 -I{} sh -c "dirname {};basename {}" |
awk '/^\// {dir=$0 ; if (dir != lastdir) {print;lastdir=dir}} !/^\// {print}'
Produces
/home/imcgowan/t/t3
jjj
iii
hhh
/home/imcgowan/t/t2
ggg
fff
eee
/home/imcgowan/t/t1
ddd
ccc
bbb
/home/imcgowan/t
aaa
while I was trying to practice my linux skills, but I could not solve this question.
So its basically saying "Write a bash script that takes a name of
directory as a command argument and printf the name of subdirectories
that has more than 5 files in it."
I thought we will use the find command but ı still could not figure it out. My code is:
find directory -type d -mindepth5
but it's not working.
You can use find twice:
First you can use find and wc to count the number of files in a given directory:
nb=$(find directory -maxdepth 1 -type f -printf "x\n" | wc -l)
This just asks find to output an x on a line for each file in the directory directory, proceeding non-recursively, then wc -l counts the number of lines, so, really, nb is the number of files in directory.
If you want to know whether a directory contains more than 5 files, it's a good idea to stop find as soon as 6 files are found:
nb=$(find directory -maxdepth 1 -type f -printf "x\n" | head -6 | wc -l)
Here nb has an upper threshold of 6.
Now if for each subdirectory of a directory directory you want to output the number of files (threshold at 6), you can do this:
find directory -type d -exec bash -c 'nb=$(find "$0" -maxdepth 1 -type f -printf "x\n" | head -6 | wc -l); echo "$nb"' {} \;
where the $0 that appears is the 0-th argument, namely {} that find will replaced by the subdirectory of directory.
Finally, you only want to display the subdirectory name if the number of files is more than 5:
find . -type d -exec bash -c 'nb=$(find "$0" -maxdepth 1 -type f -printf "x\n" | head -6 | wc -l); ((nb>5))' {} \; -print
The final test ((nb>5)) returns success or failure whether nb is greater than 5 or not, and in case of success, find will -print the subdirectory name.
This should do the trick:
find directory/ -type f | sed 's/\(.*\)\/.*/\1/g' | sort | uniq -c | sort -n | awk '{if($1>5) print($2)}'
Using mindpeth is useless here since it only lists directories at at least depth 5. You say you need subdirectories with more then 5 files in it.
find directory -type f prints all files in subdirectories
sed 's/\(.*\)\/.*/\1/g' removes names of files leaving only list of subdirecotries without filenames
sort sorts that list so we can use uniq
uniq -c merges duplicate lines and writes how many times it occured
sort -n sorts it by number of occurences (so you end up with a list:(how many times, subdirectory))
awk '{if($1>5) print($2)}' prints only those with first comlun 1 > 5 (and it only prints the second column)
So you end up with a list of subdirectories with at least 5 files inside.
EDIT:
A fix for paths with spaces was proposed:
Instead of awk '{if($1>5) print($2)}' there should be awk '{if($1>5){ $1=""; print(substr($0,2)) }}' which sets first part of line to "" and then prints whole line without a leading space (which was delimiter). So put together we get this:
find directory/ -type f | sed 's/\(.*\)\/.*/\1/g' | sort | uniq -c | sort -n | awk '{if($1>5){ $1=""; print(substr($0,2)) }}'
I am able to list all the directories by
find ./ -type d
I attempted to list the contents of each directory and count the number of files in each directory by using the following command
find ./ -type d | xargs ls -l | wc -l
But this summed the total number of lines returned by
find ./ -type d | xargs ls -l
Is there a way I can count the number of files in each directory?
This prints the file count per directory for the current directory level:
du -a | cut -d/ -f2 | sort | uniq -c | sort -nr
Assuming you have GNU find, let it find the directories and let bash do the rest:
find . -type d -print0 | while read -d '' -r dir; do
files=("$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[#]}" "$dir"
done
To check the maximum files that your system can watch:
cat /proc/sys/fs/inotify/max_user_watches
node_modules folder should be added to your IDE/editor excluded paths in slow systems, and the other files count shouldn't ideally exceed the maximum (which can be changed though).
Easy Method:
find ./|grep "Search_file.txt" |cut -d"/" -f2|sort |uniq -c
In my case I needed the count at subfolder level, so I did:
du -a | cut -d/ -f3 | sort | uniq -c | sort -nr
Easy way to recursively find files of a given type. In this case, .jpg files for all folders in current directory:
find . -name *.jpg -print | wc -l
omg why the complex commands. just use something like
find whatever_folder | wc -l
I have a Textfile with one Filename per row:
Interpret 1 - Song 1.mp3
Interpret 2 - Song 2.mp3
...
(About 200 Filenames)
Now I want to search a Folder recursivly for this Filenames to get the full path for each Filename in Filenames.txt.
How to do this? :)
(Purpose: Copied files to my MP3-Player but some of them are broken and i want to recopy them all without spending hours of researching them out of my music folder)
The easiest way may be the following:
cat orig_filenames.txt | while read file ; do find /dest/directory -name "$file" ; done > output_file_with_paths
Much faster way is run the find command only once and use fgrep.
find . -type f -print0 | fgrep -zFf ./file_with_filenames.txt | xargs -0 -J % cp % /path/to/destdir
You can use a while read loop along with find:
filecopy.sh
#!/bin/bash
while read line
do
find . -iname "$line" -exec cp '{}' /where/to/put/your/files \;
done < list_of_files.txt
Where list_of_files.txt is the list of files line by line, and /where/to/put/your/files is the location you want to copy to. You can just run it like so in the directory:
$ bash filecopy.sh
+1 for #jm666 answer, but the -J option doesn't work for my flavor of xargs, so i chaned it to:
find . -type f -print0 | fgrep -zFf ./file_with_filenames.txt | xargs -0 -I{} cp "{}" /path/to/destdir/