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 need to count all the .txt files in the current folder.
I tried ls | grep .txt but if my folder content is: a.txt btxt c.c it will select a.txt and btxt and I only want files that end with .txt. I tried various combinations of regexp but with no result.
Find may be better than in this case since it is designed for handling file names:
find . -maxdepth 0 -name '*.txt' | wc -l
Buf if you are very cautious about possibly strange file names:
find . -maxdepth 0 -name '*.txt' -exec echo 1 \; | wc -l
For Grep, using the character '.' means: "any character"... so you'll need to escape the dot:
ls | grep -e "\.txt"
edit in fact the -e option is not even necessary. this will do the trick:
ls | grep "\.txt"
If all you need is number of files with extension '.txt' in current directory only, then this will also help.
ls -l *.txt | 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
I am new to Linux shell and I found a way to get the name of the file I want:
ls *.*g -S| grep -v ^d | head -1
I am going to be repeating this for a number of file. I am trying to copy this file to another directory (cp command?). But the below code is failing.
I am trying this, but its not working:
cp ls -S| grep -v ^d | head -1 ../directory
Also, I was wondering how to loop through directorys that are in a particular directory.
Any help is much appreciated.
Thanks,
Bryan
cp $(ls *.*g -S| grep -v ^d | head -1) ../directory
cp "$(find . -type f -name "*.*" -printf "%f:%s\n" | sort -t":" -k2 -n | awk -F":" 'END{print $1}')" ../directory
Use quotes in case there are white spaces in your file. add -maxdepth 1 if you don't want to recurse subdirectories
Basically I want do the following:
ls -l[+someflags]
(or by some other means) that will only display files that are symbolic links
so the output would look
-rw-r--r-- 1 username grp size date-time filename -> somedir
-rw-r--r-- 1 username grp size date-time filename2 -> somsdfsdf
etc.
For example,
to show only directories I have an alias:
alias lsd 'ls -l | grep ^d'
I wonder how to display only hidden files or only hidden directories?
I have the following solution, however it doesn't display the output in color :(
ls -ltra | grep '\->'
Find all the symbolic links in a directory:
ls -l `find /usr/bin -maxdepth 1 -type l -print`
For the listing of hidden files:
ls -ald .*
For only "hidden" folders - dot folders, try:
ls -l .**
Yes, the two asterisks are necessary, otherwise you'll also get . and .. in the results.
For symlinks, well, try the symlinks program:
symlinks -v .
(shows all symlinks under current directory)
ls -l | grep lrw
shows only symlinks (files and directories). Not sure how to get them colorful, though.
ls -lad .*
shows only hidden files/directories
ls -l | grep drw
shows directories only.
To display JUST the symlinks and what they link to:
find -P . -type l -exec echo -n "{} -> " \; -exec readlink {} \;
To limit to JUST THIS DIR
find -P . -maxdepth 1 -type l -exec echo -n "{} -> " \; -exec readlink {} \;
Example output (after ln -s /usr/bin moo):
./moo -> /usr/bin
You were almost there with your grep solution; let's focus on getting you COLOR again.
Try this:
ls --color=always -ltra | grep '->'
Improving a little on the accepted answer given by #ChristopheD (coudnt comment on the accepted answer since I dont have enough reputation)
I use an alias
findsymlinks <path> <depth>
where the alias is
alias findsymlinks "find \!:1 -maxdepth \!:2 -type l -print | xargs ls -l --color=auto"
Try file type flag and get rid of the appending #
ls -F /home/usr/foo | grep "#" | sed 's/#//'
For (t)csh:
ls --color=always -ltra | grep '\->'
(This is simply pbr's answer but with the hyphen escaped.)
Mac OSX
On OSX, ls works differently, so add this to your ~/.cshrc file:
setenv CLICOLOR_FORCE 1 # (equivalent of Linux --color=always)
And then call:
ls -G -ltra | grep '\->' # (-G is equivalent of ls --color)
For bash:
This provides a nice output.
sl=`find -L /path/to/target -xtype l`; for links in $sl; do ls --color=always -ltra $links; done | sed 's/^/ /'
Usage: foo $path
Uses current path if none specified.
#!/bin/bash
case "$1" in
-r)
find $2 -type l -print | while IFS= read line ; do ls -l --color=always "$line"; done
;;
--help)
echo 'Usage: foo [-r] [$PATH]'
echo
echo '-r Recursive'
;;
*)
ls --color=always -ltra $1 | grep '\->'
esac