How do I display the content of files regular files matched with grep command? For example I grep a directory in order to see the regular files it has. I used the next line to see the regular files only:
ls -lR | grep ^-
Then I would like to display the content of the files found there. How do I do it?
I would do something like:
$ cat `ls -lR | egrep "^-" | rev | cut -d ' ' -f 1 | rev`
Use ls to find the files
grep finds your pattern
reverse the whole result
cut out the first file separated field to get the file name (files with spaces are problematic)
reverse the file name back to normal direction
Backticks will execute that and return the list of file names to cat.
or the way I would probably do it is use vim to look at each file.
$ vim `ls -lR | egrep "^-" | rev | cut -d ' ' -f 1 | rev`
It feels like you are trying to find only the files recursively. This is what I do in those cases:
$ vim `find . -type f -print`
There are multiple ways of doing it. Would try to give you a few easy and clean ways here. All of them handle filenames with space.
$ find . -type f -print0 | xargs -0 cat
-print0 adds a null character '\0' delimiter and you need to call xargs -0 to recognise the null delimiter. If you don't do that, whitespace in the filename create problems.
e.g. without -print0 filenames: abc 123.txt and 1.inc would be read as three separate files abc, 123.txt and 1.inc.
with -print0 this becomes abc 123.txt'\0' and 1.inc'\0' and would be read as abc 123.txt and 1.inc
As for xargs, it can accept the input as a parameter. command1 | xargs command2 means the output of command1 is passed to command2.
cat displays the content of the file.
$ find . -type f -exec echo {} \; -exec cat {} \;
This is just using the find command. It finds all the files (type f), calls echo to output the filename, then calls cat to display its content.
If you don't want the filename, omit -exec echo {} \;
Alternatively you can use cat command and pass the output of find.
$ cat `find . -type f -print`
If you want to scroll through the content of multiple files one by one. You can use.
$ less `find . -type f -print`
When using less, you can navigate through :n and :p for next and previous file respectively. press q to quit less.
I was trying to code a script that counts the number of files with a vowel in a directory.
If I use
find $1 -type f | wc -l
I get the number of files in the directory $1, but I do not know how to use grep to count just the one with a vowel, I was trying something like this
find $1 -type f | grep -l '[a,e,i,o,u,A,E,I,O,U]' | wc -l
You can use this gnu find command to count all the files with at least one vowel:
find . -maxdepth 1 -type f -iname '*[aeiou]*' -printf ".\n" | wc -l
-iname '*[aeiou]*' glob pattern will match only filename with at least one of the a,e,i,o,u (ignore case).
Remove -maxdepth 1 if you want to count files recursively in sub directories as well.
If you can accept counting directories:
ls -d *a* *e* *i* *o* *u* *y* *A* *E* *I* *O* *U* *Y* | wc -l
Otherwise:
find $1 -type f | grep -i '[aeiouy]' | wc -l
Your attempt fails for two reasons. First, -l does not make sense if grep is reading in a pipeline, since the purpose of -l is to print only the input file that matched, but in this case the only input file is stdin. Second, your syntax is wrong. Try:
... | grep -i '[aeiou]' | ...
Please don't use commas in a character group expression (the thing in [] brackets)
The best way is to first do a find(1) to get the files you want to scan. Then you need the base names, as the path info is not valid. Finally, you need to grep with [aeiouAEIOU] to get only the lines with a vowel in, and finally use wc(1) to count lines.
find ${DIRECTORY} -type f -print | sed -e 's#^.*/##' | grep '[aeiouAEIOU]' | wc -l
-type f allows you to select just files (not directories). The sed(1) command edits the output, line by line, eliminating the first part of the name up to the last / character. The grep filters names with at least one vowel and discards the others, and finally wc -l counts the lines.
while I was trying to practice my linux skills, but I could not solve this question.
So its basically saying "Write a bash script that takes a name of
directory as a command argument and printf the name of subdirectories
that has more than 5 files in it."
I thought we will use the find command but ı still could not figure it out. My code is:
find directory -type d -mindepth5
but it's not working.
You can use find twice:
First you can use find and wc to count the number of files in a given directory:
nb=$(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'm using bash.
Suppose I have a log file directory /var/myprogram/logs/.
Under this directory I have many sub-directories and sub-sub-directories that include different types of log files from my program.
I'd like to find the three newest files (modified most recently), whose name starts with 2010, under /var/myprogram/logs/, regardless of sub-directory and copy them to my home directory.
Here's what I would do manually
1. Go through each directory and do ls -lt 2010*
to see which files starting with 2010 are modified most recently.
2. Once I go through all directories, I'd know which three files are the newest. So I copy them manually to my home directory.
This is pretty tedious, so I wondered if maybe I could somehow pipe some commands together to do this in one step, preferably without using shell scripts?
I've been looking into find, ls, head, and awk that I might be able to use but haven't figured the right way to glue them together.
Let me know if I need to clarify. Thanks.
Here's how you can do it:
find -type f -name '2010*' -printf "%C#\t%P\n" |sort -r -k1,1 |head -3 |cut -f 2-
This outputs a list of files prefixed by their last change time, sorts them based on that value, takes the top 3 and removes the timestamp.
Your answers feel very complicated, how about
for FILE in find . -type d; do ls -t -1 -F $FILE | grep -v "/" | head -n3 | xargs -I{} mv {} ..; done;
or laid out nicely
for FILE in `find . -type d`;
do
ls -t -1 -F $FILE | grep -v "/" | grep "^2010" | head -n3 | xargs -I{} mv {} ~;
done;
My "shortest" answer after quickly hacking it up.
for file in $(find . -iname *.php -mtime 1 | xargs ls -l | awk '{ print $6" "$7" "$8" "$9 }' | sort | sed -n '1,3p' | awk '{ print $4 }'); do cp $file ../; done
The main command stored in $() does the following:
Find all files recursively in current directory matching (case insensitive) the name *.php and having been modified in the last 24 hours.
Pipe to ls -l, required to be able to sort by modification date, so we can have the first three
Extract the modification date and file name/path with awk
Sort these files based on datetime
With sed print only the first 3 files
With awk print only their name/path
Used in a for loop and as action copy them to the desired location.
Or use #Hasturkun's variant, which popped as a response while I was editing this post :)