"Robustly" -- I'm trying to handle files with spaces and newlines and other special characters in the file names -- only because I've seen some special characters despite being told my target environment wouldn't have them.
"Portably" -- needs to run on a wide variety of Linux machines, including BusyBox.
"Oldest" -- by modification time, with names as a tiebreaker would suffice for my purpose.
Normally I'd use find -type f -printf '%T+ %p\n' 2>/dev/null | sort | cut -d' ' -f2 | while read filename; do mything "$filename"; done
But on this particular BusyBox v1.18.4 find does not support -printf so I'm not sure I can use find to expose the modification times for sorting.
Usage: find [PATH]... [EXPRESSION]
Search for files. The default PATH is the current directory,
default EXPRESSION is '-print'
EXPRESSION may consist of:
-follow Follow symlinks
-xdev Don't descend directories on other filesystems
-maxdepth N Descend at most N levels. -maxdepth 0 applies
tests/actions to command line arguments only
-mindepth N Don't act on first N levels
-name PATTERN File name (w/o directory name) matches PATTERN
-iname PATTERN Case insensitive -name
-path PATTERN Path matches PATTERN
-regex PATTERN Path matches regex PATTERN
-type X File type is X (X is one of: f,d,l,b,c,...)
-perm NNN Permissions match any of (+NNN), all of (-NNN),
or exactly NNN
-mtime DAYS Modified time is greater than (+N), less than (-N),
or exactly N days
-mmin MINS Modified time is greater than (+N), less than (-N),
or exactly N minutes
-newer FILE Modified time is more recent than FILE's
-inum N File has inode number N
-user NAME File is owned by user NAME (numeric user ID allowed)
-group NAME File belongs to group NAME (numeric group ID allowed)
-depth Process directory name after traversing it
-size N[bck] File size is N (c:bytes,k:kbytes,b:512 bytes(def.))
+/-N: file size is bigger/smaller than N
-links N Number of links is greater than (+N), less than (-N),
or exactly N
-print Print (default and assumed)
-print0 Delimit output with null characters rather than
newlines
-exec CMD ARG ; Run CMD with all instances of {} replaced by the
matching files
-prune Stop traversing current subtree
-delete Delete files, turns on -depth option
(EXPR) Group an expression
And I read that looping over the output of ls is a bad idea.
My environment also does not support stat -c format but I can use stat -t ...
So my current idea is to
cd "$1"
if [ $? -ne 0 ]; then
printf 'failed cd'
exit 1
fi
while read name
do
mything "$name"
done <<< "$(stat -t ./* \
| awk -F' ' '{print $13 " " $1}' \
| sort -r \
| awk -F' ' '{print $2}')"
Is there a better way? I'm relying on stat's 'terse' format being consistent.
Related
How can use the ls command and options to list the repetitious filenames that are in different directories?
You can't use a single, basic ls command to do this. You'd have to use a combination of other POSIX/Unix/GNU utilities. For example, to find the duplicate filenames first:
find . -type f -exec basename "\{}" \; | sort | uniq -d > dupes
This means find all the files (-type f) through the entire directory hierarchy in the current directory (.), and execute (-exec) the command basename (which strips the directory portion) on the found file (\{}), end of command (\;). These files then sort and print out duplicate lines (uniq -d). The result goes in the file dupes. Now you have the filenames that are duplicated, but you don't know what directory they are in. Use find again to find them. Using bash as your shell:
while read filename; do find . -name "$filename" -print; done < dupes
This means loop through (while) all contents of file dupes and read into the variable filename each line. For each line, execute find again and search for the specific -name of the $filename and print it out (-print, but it's implicit so this is redundant).
Truth be told you can combine these without using an intermediate file:
find . -type f -exec basename "\{}" \; | sort | uniq -d | while read filename; do find . -name "$filename" -print; done
If you're not familiar with it, the | operator means, execute the following command using the output of the previous command as the input to the following command. Example:
eje#EEWANCO-PC:~$ mkdir test
eje#EEWANCO-PC:~$ cd test
eje#EEWANCO-PC:~/test$ mkdir 1 2 3 4 5
eje#EEWANCO-PC:~/test$ mkdir 1/2 2/3
eje#EEWANCO-PC:~/test$ touch 1/0000 2/1111 3/2222 4/2222 5/0000 1/2/1111 2/3/4444
eje#EEWANCO-PC:~/test$ find . -type f -exec basename "\{}" \; | sort | uniq -d | while read filename; do find . -name "$filename" -print; done
./1/0000
./5/0000
./1/2/1111
./2/1111
./3/2222
./4/2222
Disclaimer: The requirement stated that the filenames were all numbers. While I have tried to design the code to handle filenames with spaces (and in tests on my system, it works), the code may break when it encounters special characters, newlines, nuls, or other unusual situations. Please note that the -exec parameter has special security considerations and should not be used by root over arbitrary user files. The simplified example provided is intended for illustrative and didactic purposes only. Please consult your man pages and relevant CERT advisories for full security implications.
I have a function in my bash profile (bash 4.4) for duplicate files.
It is true that find is the correct tool.
I use find combined with -print0 options which separates the find results with null char instead of new lines (default find action). Now i can catch all files under current directory and subdirectories.
This will ensure that results will be correct no matter if filenames contain special chars like spaces or new lines (in some very rare cases). Instead of double running find against find, you can built an array and just locate the duplicate files in this array. Then you grep the whole array using the "duplicates" as pattern.
So something like this works ok for my function:
$ IFS= readarray -t -d '' fn< <(find . -name 'file*' -print0)
$ dupes=$(LC_ALL=C sort <(printf '\<%s\>$\n' "${fn[#]##*/}") |uniq -d)
$ grep -e "$dupes" <(printf '%s\n' "${fn[#]}") |awk -F/ '{print $NF,"==>",$0}' |LC_ALL=C sort
This is a test:
$ IFS= readarray -t -d '' fn< <(find . -name 'file*' -print0)
# find all files and load them in an array using null delimiter
$ printf '%s\n' "${fn[#]}" #print the array
./tmp/file7
./tmp/file14
./tmp/file11
./tmp/file8
./tmp/file9
./tmp/tmp2/file09 99
./tmp/tmp2/file14.txt
./tmp/tmp2/file15.txt
./tmp/tmp2/file$100
./tmp/tmp2/file14.txt.bak
./tmp/tmp2/file15.txt.bak
./tmp/file1
./tmp/file4
./file09 99
./file14
./file$100
./file1
$ dupes=$(LC_ALL=C sort <(printf '\<%s\>$\n' "${fn[#]##*/}") |uniq -d)
#Locate duplicate files
$ echo "$dupes"
\<file$100\>$ #Mind this one with special char $ in filename
\<file09 99\>$ #Mind also this one with spaces
\<file14\>$
\<file1\>$
#I have on purpose enclose the results between \<...\> to force grep later to capture full words and avoid file1 to match file1.txt or file11
$ grep -e "$dupes" <(printf '%s\n' "${fn[#]}") |awk -F/ '{print $NF,"==>",$0}' |LC_ALL=C sort
file$100 ==> ./file$100 #File with special char correctly captured
file$100 ==> ./tmp/tmp2/file$100
file09 99 ==> ./file09 99 #File with spaces in name also correctly captured
file09 99 ==> ./tmp/tmp2/file09 99
file1 ==> ./file1
file1 ==> ./tmp/file1
file14 ==> ./file14 #other files named file14 like file14.txt and file14.txt.bak not captured since they are not duplicates.
file14 ==> ./tmp/file14
Tips:
This one <(printf '\<%s\>$\n' "${fn[#]##*/}") uses process substitution on the basename of the find results using bash built in parameter expansion techniques.
LC_ALL=C is required on sorting in order filenames to be sorted correctly.
In bash versions before 4.4 , the readarray does not accept -d option (delimiter). In this case you can transform find results to an array with
while IFS= read -r -d '' res;do fn+=( "$res" );done < <(find.... -print0)
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[#]}" "$dir"
done
To check the maximum files that your system can watch:
cat /proc/sys/fs/inotify/max_user_watches
node_modules folder should be added to your IDE/editor excluded paths in slow systems, and the other files count shouldn't ideally exceed the maximum (which can be changed though).
Easy Method:
find ./|grep "Search_file.txt" |cut -d"/" -f2|sort |uniq -c
In my case I needed the count at subfolder level, so I did:
du -a | cut -d/ -f3 | sort | uniq -c | sort -nr
Easy way to recursively find files of a given type. In this case, .jpg files for all folders in current directory:
find . -name *.jpg -print | wc -l
omg why the complex commands. just use something like
find whatever_folder | wc -l
Using Gnome in Linux Mint 12, I copied a Folder of about 9.7 GB (containing a complex tree of subfolders) from one NTFS Flash Drive to another NTFS Flash Drive. According to Gnome the file counts match, but according to du (and other programs) the byte counts don't match. (I've had the same problem copying folders in other Linux distros and Windows XP.)
I only want to know which files don't have matching byte counts. (I don't want to compare the contents of each file, because that would take way too long.) What's the best, easiest and fastest way to find the byte-count-mismatched files?
I would adapt the answer by #user1464130 as it has trouble handling spaces in file names.
cd dir1
find . -type f -printf "%p %s\n" | sort > ~/dir1.txt
cd dir2
find . -type f -printf "%p %s\n" | sort > ~/dir2.txt
diff ~/dir1.txt ~/dir2.txt
If you want to launch a command on each file and use the result in the report, you can use the while Bash construct. This example uses md5sum to compute a checksum for each file.
find . -maxdepth 1 -type f -printf "%p %s\n" | while read path size; do echo "$path - $(md5sum $path | tr -s " " | cut -f 1 -d " ") - $size" ; done
Each $() is executed separately and allows us to compute the checksum for each file. The use of tr squeezes every consecutive spaces into a single space and cut extracts the word in the n-th position, here in the first position. If we don't do that, we get the name of the file two times because md5sum give it back on stdout.
Here is an example without using the comparison (no diff). Note that I've used a dash - to emphasize the three datas we output about each file but it could be a problem if you want to feed it to another program.
$ find . -maxdepth 1 -name "*.c" -type f -printf "%p %s\n" | while read path size; do echo "$path - $(md5sum $path | tr -s " " | cut -f 1 -d " ") - $size" ; done
./thread.c - 5f2b7b12c7cd12fcb9e9796078e5d15b - 584
./utils.c - d61bc1dbc72768e622a04f03e3b8f7a2 - 3413
EDIT : And to handle spaces in filenames and still get the checksum and the size, you can use the following code.
$ find . -maxdepth 1 -name "*.c" -type f -print0 | xargs -0 -n 1 md5sum | while read checksum path; do echo $path $(stat --printf="%s" "$path") $checksum ; done
./ini tia li za tion.c 84 31626123e9056bac2e96b472bd62f309
Did you check if both partitions have the same attributes? (block size, size, reserved space for deletions or bad blocks, etc.)
For your specific case, I would recommend rsync with option -n (or --dry-run). It will tell you which files are different. That is:
$ rsync -I -n /source/ /target/
The option -I is to ignore times. You can use the same command to make both directories equivalent (timestamp, permissions, etc.).
Check the manual of rsync or try the option --help to get more options and examples on how to use it. It is very powerful.
Assuming you need to compare dir1 and dir 2, here are the console commands:
cd dir1
find . -type f|sort|xargs ls -l| awk '{print $5,$8}' > ~/dir1.txt
cd dir2
find . -type f|sort|xargs ls -l| awk '{print $5,$8}' > ~/dir2.txt
diff ~/dir1.txt ~/dir2.txt
You may need to edit awk parameters to make it print file length and path properly.