ls /etc | tail -1 | wc -l
so basically I used this command but it counts the number of files that I've got from the tail command (which is the last file in the directory=1) but I didn't get the number of lines that are in the file.
and I used the cat command to open the file and count the lines but it didn't work.
ls /etc | cat tail -1 | wc -l
ls /etc | tail -1 | cat |wc -l
You could use xargs to use the result of the tail as an argument for wc, although I'd recommend using find instead of ls so you get the full path and don't need mess around with relative pathes:
$ find /etc -type f | tail -1 | xargs wc -l
You should never parse ls (instead parse /etc/*)
$ wc -l < `find /etc -maxdepth 1 -type f | tail -n 1`
or
$ find /etc -maxdepth 1 -type f | tail -n 1 | wc -l
What this does is find the last file for /etc
And puts it's content in wc -l
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
How can I display all files greater than 10k bytes in my current directory and it's subdirectories.
Tried ls -size +10k but that didn't work.
find . -size +10k -exec ls -lh {} \+
the first part of this is identical to #sputnicks answer, and sucesffully finds all files in the directory over 10k (don't confuse k with K), my addition, the second part then executes ls -lh or ls that lists(-l) the files by human readable size(-h). negate the h if you prefer. of course the {} is the file itself, and the \+ is simply an alternative to \;
which in practice \; would repeat or:
ls -l found.file; ls -l found.file.2; ls -l found.file.3
where \+ display it as one statement or:
ls -l found.file found.file.2 found.file.3
more on \; vs + with find
Additionaly, you may want the listing ordered by size. Which is relatively easy to accomplish. I would at the -s option to ls, so ls -ls and then pipe it to sort -n to sort numerically
which would become:
find . -size +10k -exec ls -ls {} \+ | sort -n
or in reverse order add an -r :
find . -size +10k -exec ls -ls {} \+ | sort -nr
finally, your title says find biggest file in directory. You can do that by then piping the code to tail
find . -size +10k -exec ls -ls {} \+ | sort -n | tail -1
would find you the largest file in the directory and its sub directories.
note you could also sort files by size by using -S, and negate the need for sort. but to find the largest file you would need to use head so
find . -size +10k -exec ls -lS {} \+ | head -1
the benefit of doing it with -S and not sort is one, you don't have to type sort -n and two you can also use -h the human readable size option. which is one of my favorite to use, but is not available with older versisions of ls, for example we have an old centOs 4 server at work that doesn't have -h
Try doing this:
find . -size +10k -ls
And if you want to use the binary ls :
find . -size +10k -exec ls -l {} \;
I realize the assignment is likely long over. For anyone else:
You are overcomplicating.
find . -size +10k
I'll add to #matchew answer (not enough karma points to comment):
find . -size +10k -type f -maxdepth 1 -exec ls -lh {} \; > myLogFile.txt
-type f :specify regular file type
-maxdepth 1 :make sure it only find files in the current directory
You may use ls like that:
ls -lR | egrep -v '^d' | awk '$5>10240{print}'
Explanation:
ls -lR # list recursivly
egrep -v '^d' # only print lines which do not start with a 'd'. (files)
only print lines where the fifth column (size) is greater that 10240 bytes:
awk '$5>10240{print}'
I'm trying to get my terminal to return the latest .txt file, with path intact. I've been researching ls, grep, find, and tail, using the '|' functionality of passing results from one utility to the next. The end result would be to have a working path + result that I could pass my text editor.
I've been getting close with tests like this:
find . | grep '.txt$' | tail -1
..but I haven't had luck with grep returning the newest file - is there a flag I'm missing?
Trying to use find & ls isn't exactly working either:
find . -name "*.txt" | ls -lrth
..the ls returns the current directories instead of the results of my find query.
Please help!
You're so very close.
vi "$(find . -name '*.txt' -exec ls -t {} + | head -1)"
find /usr/share -name '*.txt' -printf '%C+ %p\n' | sort -r | head -1 | sed 's/^[^ ]* //'
If you have bash4+
ls -t ./**/*.txt | head -1
edit the latest txt file
vim $(ls -t ./**/*.txt |head -1)
ps: need enabled shopt -s globstar in your .bashrc or .profile...
You can use the stat function to print each file with just the latest modification time and name.
find . -name "*.txt" -exec stat -c "%m %N" {} \; | sort