I am new to shell scripting, so I need some help here. I have a directory that fills up with backups. If I have more than 10 backup files, I would like to remove the oldest files, so that the 10 newest backup files are the only ones that are left.
So far, I know how to count the files, which seems easy enough, but how do I then remove the oldest files, if the count is over 10?
if [ls /backups | wc -l > 10]
then
echo "More than 10"
fi
Try this:
ls -t | sed -e '1,10d' | xargs -d '\n' rm
This should handle all characters (except newlines) in a file name.
What's going on here?
ls -t lists all files in the current directory in decreasing order of modification time. Ie, the most recently modified files are first, one file name per line.
sed -e '1,10d' deletes the first 10 lines, ie, the 10 newest files. I use this instead of tail because I can never remember whether I need tail -n +10 or tail -n +11.
xargs -d '\n' rm collects each input line (without the terminating newline) and passes each line as an argument to rm.
As with anything of this sort, please experiment in a safe place.
find is the common tool for this kind of task :
find ./my_dir -mtime +10 -type f -delete
EXPLANATIONS
./my_dir your directory (replace with your own)
-mtime +10 older than 10 days
-type f only files
-delete no surprise. Remove it to test your find filter before executing the whole command
And take care that ./my_dir exists to avoid bad surprises !
Make sure your pwd is the correct directory to delete the files then(assuming only regular characters in the filename):
ls -A1t | tail -n +11 | xargs rm
keeps the newest 10 files. I use this with camera program 'motion' to keep the most recent frame grab files. Thanks to all proceeding answers because you showed me how to do it.
The proper way to do this type of thing is with logrotate.
I like the answers from #Dennis Williamson and #Dale Hagglund. (+1 to each)
Here's another way to do it using find (with the -newer test) that is similar to what you started with.
This was done in bash on cygwin...
if [[ $(ls /backups | wc -l) > 10 ]]
then
find /backups ! -newer $(ls -t | sed '11!d') -exec rm {} \;
fi
Straightforward file counter:
max=12
n=0
ls -1t *.dat |
while read file; do
n=$((n+1))
if [[ $n -gt $max ]]; then
rm -f "$file"
fi
done
I just found this topic and the solution from mikecolley helped me in a first step. As I needed a solution for a single line homematic (raspberrymatic) script, I ran into a problem that this command only gave me the fileames and not the whole path which is needed for "rm". My used CUxD Exec command can not start in a selected folder.
So here is my solution:
ls -A1t $(find /media/usb0/backup/ -type f -name homematic-raspi*.sbk) | tail -n +11 | xargs rm
Explaining:
find /media/usb0/backup/ -type f -name homematic-raspi*.sbk searching only files -type f whiche are named like -name homematic-raspi*.sbk (case sensitive) or use -iname (case insensitive) in folder /media/usb0/backup/
ls -A1t $(...) list the files given by find without files starting with "." or ".." -A sorted by mtime -t and with a return of only one column -1
tail -n +11 return of only the last 10 -n +11 lines for following rm
xargs rm and finally remove the raiming files in the list
Maybe this helps others from longer searching and makes the solution more flexible.
stat -c "%Y %n" * | sort -rn | head -n +10 | \
cut -d ' ' -f 1 --complement | xargs -d '\n' rm
Breakdown: Get last-modified times for each file (in the format "time filename"), sort them from oldest to newest, keep all but the last ten entries, and then keep all but the first field (keep only the filename portion).
Edit: Using cut instead of awk since the latter is not always available
Edit 2: Now handles filenames with spaces
On a very limited chroot environment, we had only a couple of programs available to achieve what was initially asked. We solved it that way:
MIN_FILES=5
FILE_COUNT=$(ls -l | grep -c ^d )
if [ $MIN_FILES -lt $FILE_COUNT ]; then
while [ $MIN_FILES -lt $FILE_COUNT ]; do
FILE_COUNT=$[$FILE_COUNT-1]
FILE_TO_DEL=$(ls -t | tail -n1)
# be careful with this one
rm -rf "$FILE_TO_DEL"
done
fi
Explanation:
FILE_COUNT=$(ls -l | grep -c ^d ) counts all files in the current folder. Instead of grep we could use also wc -l but wc was not installed on that host.
FILE_COUNT=$[$FILE_COUNT-1] update the current $FILE_COUNT
FILE_TO_DEL=$(ls -t | tail -n1) Save the oldest file name in the $FILE_TO_DEL variable. tail -n1 returns the last element in the list.
Based on others suggestions and some awk foo, I got this to work. I know this an old thread, but I didn't find a decent answer here and this sorted it for me. This just deletes the oldest file, but you can change the head -n 1 to 10 and get the oldest 10.
find $DIR -type f -printf '%T+ %p\n' | sort | head -n 1 | awk '{first =$1; $1 =""; print $0}' | xargs -d '\n' rm
Using inode numbers via stat & find command (to avoid pesky-chars-in-file-name issues):
stat -f "%m %i" * | sort -rn -k 1,1 | tail -n +11 | cut -d " " -f 2 | \
xargs -n 1 -I '{}' find "$(pwd)" -type f -inum '{}' -print
#stat -f "%m %i" * | sort -rn -k 1,1 | tail -n +11 | cut -d " " -f 2 | \
# xargs -n 1 -I '{}' find "$(pwd)" -type f -inum '{}' -delete
Related
I would like to find the newest sub directory in a directory and save the result to variable in bash.
Something like this:
ls -t /backups | head -1 > $BACKUPDIR
Can anyone help?
BACKUPDIR=$(ls -td /backups/*/ | head -1)
$(...) evaluates the statement in a subshell and returns the output.
There is a simple solution to this using only ls:
BACKUPDIR=$(ls -td /backups/*/ | head -1)
-t orders by time (latest first)
-d only lists items from this folder
*/ only lists directories
head -1 returns the first item
I didn't know about */ until I found Listing only directories using ls in bash: An examination.
This ia a pure Bash solution:
topdir=/backups
BACKUPDIR=
# Handle subdirectories beginning with '.', and empty $topdir
shopt -s dotglob nullglob
for file in "$topdir"/* ; do
[[ -L $file || ! -d $file ]] && continue
[[ -z $BACKUPDIR || $file -nt $BACKUPDIR ]] && BACKUPDIR=$file
done
printf 'BACKUPDIR=%q\n' "$BACKUPDIR"
It skips symlinks, including symlinks to directories, which may or may not be the right thing to do. It skips other non-directories. It handles directories whose names contain any characters, including newlines and leading dots.
Well, I think this solution is the most efficient:
path="/my/dir/structure/*"
backupdir=$(find $path -type d -prune | tail -n 1)
Explanation why this is a little better:
We do not need sub-shells (aside from the one for getting the result into the bash variable).
We do not need a useless -exec ls -d at the end of the find command, it already prints the directory listing.
We can easily alter this, e.g. to exclude certain patterns. For example, if you want the second newest directory, because backup files are first written to a tmp dir in the same path:
backupdir=$(find $path -type -d -prune -not -name "*temp_dir" | tail -n 1)
The above solution doesn't take into account things like files being written and removed from the directory resulting in the upper directory being returned instead of the newest subdirectory.
The other issue is that this solution assumes that the directory only contains other directories and not files being written.
Let's say I create a file called "test.txt" and then run this command again:
echo "test" > test.txt
ls -t /backups | head -1
test.txt
The result is test.txt showing up instead of the last modified directory.
The proposed solution "works" but only in the best case scenario.
Assuming you have a maximum of 1 directory depth, a better solution is to use:
find /backups/* -type d -prune -exec ls -d {} \; |tail -1
Just swap the "/backups/" portion for your actual path.
If you want to avoid showing an absolute path in a bash script, you could always use something like this:
LOCALPATH=/backups
DIRECTORY=$(cd $LOCALPATH; find * -type d -prune -exec ls -d {} \; |tail -1)
With GNU find you can get list of directories with modification timestamps, sort that list and output the newest:
find . -mindepth 1 -maxdepth 1 -type d -printf "%T#\t%p\0" | sort -z -n | cut -z -f2- | tail -z -n1
or newline separated
find . -mindepth 1 -maxdepth 1 -type d -printf "%T#\t%p\n" | sort -n | cut -f2- | tail -n1
With POSIX find (that does not have -printf) you may, if you have it, run stat to get file modification timestamp:
find . -mindepth 1 -maxdepth 1 -type d -exec stat -c '%Y %n' {} \; | sort -n | cut -d' ' -f2- | tail -n1
Without stat a pure shell solution may be used by replacing [[ bash extension with [ as in this answer.
Your "something like this" was almost a hit:
BACKUPDIR=$(ls -t ./backups | head -1)
Combining what you wrote with what I have learned solved my problem too. Thank you for rising this question.
Note: I run the line above from GitBash within Windows environment in file called ./something.bash.
I'm trying to delete the first largest regular files from the given directory, but it doesn't work for files which contain whitespace caracters.
My code (it works if the files doesn't contain whitespace caracters):
find mydir -type f -exec du -ahb {} + | sort -n -r | cut -f2 | head -n 10 | xargs rm -i
I also tried this, but it gives an error message:
find mydir -type f -exec du -ahb {} + -print 0 | sort -n -r | cut -f2 | head -n 10 | xargs -0 rm -i
The following should work at least with GNU coreutils 8.25 and newer :
find mydir -type f -exec du -0b {} + | sort -znr | cut -zf2 | head -zn 10 | xargs -0pn 1 rm
I made sure every command handled and outputted NUL bytes (\0) separated records rather than linefeed separated records :
du outputs NUL-separated records with -0
sort, cut and head handle and output NUL-separated records with -z
xargs handles NUL-separated records with -0
Additionally, I removed the interactive mode of rm and asked xargs to handle that instead (-p), because xargs didn't provide a prompt to rm when invoking it. I had to limit the number of parameters given at once to rm to 1 for this to work (xargs' -n 1 parameter). There might be a way to preserve the -i and provide rm with an interface to your prompt, but I don't know how.
Last point : I removed du's -human-readable mode because it would have made the sort often fail and it didn't serve any purpose since the filesizes were never displayed to an human.
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 using Linux and would like to replace all files containing the string 000000 with an existing file /home/user/offblack.png but keep the existing filename. I've been working at this for a while with various combinations of -exec and xargs but no luck. So far I have:
find | grep 000000
Which does list all the files I want to change fine. How do I copy and replace these files with my existing offblack.png file?
Here's what I would use:
find (your find args here) \
| xargs fgrep '000000' /dev/null \
| awk -F: '{print $1}' \
| xargs -n 1 -I ORIGINAL_FILENAME /bin/echo /bin/cp /path/to/offblack.png ORIGINAL_FILENAME
Expanding, find all the files you're interested in, grep inside of them for the string '000000' (adding /dev/null to the list of files in case one of the generated fgreps ended up with only one filename - it ensures the output is always formatted as "filename: <line containing '000000'>"), strip out only the filenames, then one-by-one, copy in offblack.png over those files. Note that I inserted a /bin/echo in there. That's your dry-run. Remove the echo to get it to run for real.
If what you mean is that the filenames contain "000000":
find . -type f -a -name '*000000*' -exec /bin/echo /bin/cp /path/to/offblack.png {} \;
Much simpler. :-) Find every file under the current directory with a name containing your string and exec the copy of offblack.png over it. Again, what I've given you there is a dry-run. Remove the echo for your live fire drill. :-)
find . -type f | grep 000000 | tr \\n \\0 | xargs -0i+ cp ~/offblack.png "+"
Let's try and use Bash a bit more:
for read -r filename
do
hit=""
for read -r
do
if [[ $REPLY == *000000* ]]
then
hit=$filename
break
fi
done < $filename
[[ -n $hit ]] && cp /path/offblack.png $filename
done < <(find . -type -f)
Fewer man pages to search!
After a few searches from Google, what I come up with is:
find my_folder -type f -exec grep -l "needle text" {} \; -exec file {} \; | grep text
which is very unhandy and outputs unneeded texts such as mime type information. Any better solutions? I have lots of images and other binary files in the same folder with a lot of text files that I need to search through.
I know this is an old thread, but I stumbled across it and thought I'd share my method which I have found to be a very fast way to use find to find only non-binary files:
find . -type f -exec grep -Iq . {} \; -print
The -I option to grep tells it to immediately ignore binary files and the . option along with the -q will make it immediately match text files so it goes very fast. You can change the -print to a -print0 for piping into an xargs -0 or something if you are concerned about spaces (thanks for the tip, #lucas.werkmeister!)
Also the first dot is only necessary for certain BSD versions of find such as on OS X, but it doesn't hurt anything just having it there all the time if you want to put this in an alias or something.
EDIT: As #ruslan correctly pointed out, the -and can be omitted since it is implied.
Based on this SO question :
grep -rIl "needle text" my_folder
Why is it unhandy? If you need to use it often, and don't want to type it every time just define a bash function for it:
function findTextInAsciiFiles {
# usage: findTextInAsciiFiles DIRECTORY NEEDLE_TEXT
find "$1" -type f -exec grep -l "$2" {} \; -exec file {} \; | grep text
}
put it in your .bashrc and then just run:
findTextInAsciiFiles your_folder "needle text"
whenever you want.
EDIT to reflect OP's edit:
if you want to cut out mime informations you could just add a further stage to the pipeline that filters out mime informations. This should do the trick, by taking only what comes before :: cut -d':' -f1:
function findTextInAsciiFiles {
# usage: findTextInAsciiFiles DIRECTORY NEEDLE_TEXT
find "$1" -type f -exec grep -l "$2" {} \; -exec file {} \; | grep text | cut -d ':' -f1
}
find . -type f -print0 | xargs -0 file | grep -P text | cut -d: -f1 | xargs grep -Pil "search"
This is unfortunately not space save. Putting this into bash script makes it a bit easier.
This is space safe:
#!/bin/bash
#if [ ! "$1" ] ; then
echo "Usage: $0 <search>";
exit
fi
find . -type f -print0 \
| xargs -0 file \
| grep -P text \
| cut -d: -f1 \
| xargs -i% grep -Pil "$1" "%"
Another way of doing this:
# find . |xargs file {} \; |grep "ASCII text"
If you want empty files too:
# find . |xargs file {} \; |egrep "ASCII text|empty"
How about this:
$ grep -rl "needle text" my_folder | tr '\n' '\0' | xargs -r -0 file | grep -e ':[^:]*text[^:]*$' | grep -v -e 'executable'
If you want the filenames without the file types, just add a final sed filter.
$ grep -rl "needle text" my_folder | tr '\n' '\0' | xargs -r -0 file | grep -e ':[^:]*text[^:]*$' | grep -v -e 'executable' | sed 's|:[^:]*$||'
You can filter-out unneeded file types by adding more -e 'type' options to the last grep command.
EDIT:
If your xargs version supports the -d option, the commands above become simpler:
$ grep -rl "needle text" my_folder | xargs -d '\n' -r file | grep -e ':[^:]*text[^:]*$' | grep -v -e 'executable' | sed 's|:[^:]*$||'
Here's how I've done it ...
1 . make a small script to test if a file is plain text
istext:
#!/bin/bash
[[ "$(file -bi $1)" == *"file"* ]]
2 . use find as before
find . -type f -exec istext {} \; -exec grep -nHi mystring {} \;
Here's a simplified version with extended explanation for beginners like me who are trying to learn how to put more than one command in one line.
If you were to write out the problem in steps, it would look like this:
// For every file in this directory
// Check the filetype
// If it's an ASCII file, then print out the filename
To achieve this, we can use three UNIX commands: find, file, and grep.
find will check every file in the directory.
file will give us the filetype. In our case, we're looking for a return of 'ASCII text'
grep will look for the keyword 'ASCII' in the output from file
So how can we string these together in a single line? There are multiple ways to do it, but I find that doing it in order of our pseudo-code makes the most sense (especially to a beginner like me).
find ./ -exec file {} ";" | grep 'ASCII'
Looks complicated, but not bad when we break it down:
find ./ = look through every file in this directory. The find command prints out the filename of any file that matches the 'expression', or whatever comes after the path, which in our case is the current directory or ./
The most important thing to understand is that everything after that first bit is going to be evaluated as either True or False. If True, the file name will get printed out. If not, then the command moves on.
-exec = this flag is an option within the find command that allows us to use the result of some other command as the search expression. It's like calling a function within a function.
file {} = the command being called inside of find. The file command returns a string that tells you the filetype of a file. Regularly, it would look like this: file mytextfile.txt. In our case, we want it to use whatever file is being looked at by the find command, so we put in the curly braces {} to act as an empty variable, or parameter. In other words, we're just asking for the system to output a string for every file in the directory.
";" = this is required by find and is the punctuation mark at the end of our -exec command. See the manual for 'find' for more explanation if you need it by running man find.
| grep 'ASCII' = | is a pipe. Pipe take the output of whatever is on the left and uses it as input to whatever is on the right. It takes the output of the find command (a string that is the filetype of a single file) and tests it to see if it contains the string 'ASCII'. If it does, it returns true.
NOW, the expression to the right of find ./ will return true when the grep command returns true. Voila.
I have two issues with histumness' answer:
It only list text files. It does not actually search them as
requested. To actually search, use
find . -type f -exec grep -Iq . {} \; -and -print0 | xargs -0 grep "needle text"
It spawns a grep process for every file, which is very slow. A better solution is then
find . -type f -print0 | xargs -0 grep -IZl . | xargs -0 grep "needle text"
or simply
find . -type f -print0 | xargs -0 grep -I "needle text"
This only takes 0.2s compared to 4s for solution above (2.5GB data / 7700 files), i.e. 20x faster.
Also, nobody cited ag, the Silver Searcher or ack-grep¸as alternatives. If one of these are available, they are much better alternatives:
ag -t "needle text" # Much faster than ack
ack -t "needle text" # or ack-grep
As a last note, beware of false positives (binary files taken as text files). I already had false positive using either grep/ag/ack, so better list the matched files first before editing the files.
Although it is an old question, I think this info bellow will add to the quality of the answers here.
When ignoring files with the executable bit set, I just use this command:
find . ! -perm -111
To keep it from recursively enter into other directories:
find . -maxdepth 1 ! -perm -111
No need for pipes to mix lots of commands, just the powerful plain find command.
Disclaimer: it is not exactly what OP asked, because it doesn't check if the file is binary or not. It will, for example, filter out bash script files, that are text themselves but have the executable bit set.
That said, I hope this is useful to anyone.
I do it this way:
1) since there're too many files (~30k) to search thru, I generate the text file list daily for use via crontab using below command:
find /to/src/folder -type f -exec file {} \; | grep text | cut -d: -f1 > ~/.src_list &
2) create a function in .bashrc:
findex() {
cat ~/.src_list | xargs grep "$*" 2>/dev/null
}
Then I can use below command to do the search:
findex "needle text"
HTH:)
I prefer xargs
find . -type f | xargs grep -I "needle text"
if your filenames are weird look up using the -0 options:
find . -type f -print0 | xargs -0 grep -I "needle text"
bash example to serach text "eth0" in /etc in all text/ascii files
grep eth0 $(find /etc/ -type f -exec file {} \; | egrep -i "text|ascii" | cut -d ':' -f1)
If you are interested in finding any file type by their magic bytes using the awesome file utility combined with power of find, this can come in handy:
$ # Let's make some test files
$ mkdir ASCII-finder
$ cd ASCII-finder
$ dd if=/dev/urandom of=binary.file bs=1M count=1
1+0 records in
1+0 records out
1048576 bytes (1.0 MB, 1.0 MiB) copied, 0.009023 s, 116 MB/s
$ file binary.file
binary.file: data
$ echo 123 > text.txt
$ # Let the magic begin
$ find -type f -print0 | \
xargs -0 -I ## bash -c 'file "$#" | grep ASCII &>/dev/null && echo "file is ASCII: $#"' -- ##
Output:
file is ASCII: ./text.txt
Legend: $ is the interactive shell prompt where we enter our commands
You can modify the part after && to call some other script or do some other stuff inline as well, i.e. if that file contains given string, cat the entire file or look for a secondary string in it.
Explanation:
find items that are files
Make xargs feed each item as a line into one liner bash
command/script
file checks type of file by magic byte, grep checks if ASCII
exists, if so, then after && your next command executes.
find prints results null separated, this is good to escape
filenames with spaces and meta-characters in it.
xargs , using -0 option, reads them null separated, -I ##
takes each record and uses as positional parameter/args to bash
script.
-- for bash ensures whatever comes after it is an argument even
if it starts with - like -c which could otherwise be interpreted
as bash option
If you need to find types other than ASCII, simply replace grep ASCII with other type, like grep "PDF document, version 1.4"
find . -type f | xargs file | grep "ASCII text" | awk -F: '{print $1}'
Use find command to list all files, use file command to verify they are text (not tar,key), finally use awk command to filter and print the result.
How about this
find . -type f|xargs grep "needle text"