I have a list of files inside directory1 and I want to move n number of files from that directory1 to directory2. When I try xargs like this it did not work.
ls -ltr | head -20 | mv xargs /directory2
Why we can't use xargs in middle? how to move n number of files to another directory in command line ?
First, you should not use ls in scripts. ls is for humans, not for scripting. Second, the last command of your pipe tries to move a file named xargs to /directory2. Third, xargs appends its inputs to the command. Even if you swap mv and xargs this will lead to execute mv /directory2 file1 file2 file3... file20, instead of what you want: mv file1 file2 file3... file20 /directory2. Finally, the -l option of ls will print more than just the file name (permissions, owner, group...); you cannot use it as mv argument.
Your ls options suggest that you want to move the 20 oldest files. Try:
while read -d '' -r time file; do
mv -- "$file" "/directory2"
done < <( stat --printf '%Y %n\0' * | sort -zn | head -zn20 )
stat --printf '%Y %n\0' * prints the last modification time as seconds since epoch, followed by the file name, of all files in the current directory. Each record is terminated by the NUL character, the only character that cannot be found in a file name. The -z option of sort and head and the -d '' option of read instruct these utilities to use NUL as the record separator instead of the default (newline). This way, the script should work even if some of your file names contain newlines.
If you prefer xargs:
stat --printf '%Y\t%n\0' * | sort -zn | head -zn20 | cut -zf2- |
xargs -0I{} mv -- {} /directory2
You can try this:
n=20
cd /path/to/directory1 || exit
files=(*)
mv -- "${files[#]:0:n}" /path/to/directory2
Also, you may consider reading this article: Why you shouldn't parse the output of ls
Line1: .................
Line2: #hello1 #hello2 #hello3
Line3: .................
Line4: .................
Line5: #hello1 #hello4 #hello3
Line6: #hello1 #hello2 #hello3
Line7: .................
I have files that look similar in terms of lines on one of my project directories. I want to get the counts of all the lines that contain #hello1 and #hello2. In this case I would get 2 as a result only for this file. However, I want to do this recursively.
The canonical way to "do something recursively" is to use the find command. If you want to find lines that have two words on them, a simple regex will do:
grep -lr '#hello1.*#hello2' .
The option -l instructs grep to show us only filenames rather than file content, and the option -r tells grep to traverse the filesystem recursively. The start of the search is the path at the end of the line. Once you have the list of files, you can parse that list using commands run by xargs.
For example, this will count all the lines in files matching the pattern you specified.
grep -lr '#hello1.*#hello2' . | xargs -n 1 wc -l
This uses xargs to run the wc command on each of the files listed by grep. You could probably also run this without the -n 1, unless you're dealing with many many thousands of files that would exceed your maximum command line length.
Or, if I'm interpreting your question correctly, the following will count just the patterns in those files.
grep -lr '#hello1.*#hello2' . | xargs -n 1 grep -Hc '#hello1.*#hello2'
This runs a similar grep to the one used to generate your recursive list of files, and presents the output with filename (-H) and count (-c).
But if you want complex rules like finding two patterns possibly on different lines in the file, then grep probably is not the optimal tool, unless you use multiple greps launched by find:
find /path/to/base -type f \
-exec grep -q '#hello1' {} \; \
-exec grep -q '#hello2' {} \; \
-print
(Lines split for easier reading.)
This is somewhat costly, as find needs to launch up to two children for each file. So another approach would be to use awk instead:
find /path/to/base -type f \
-exec awk '/#hello1/{c++} /#hello2/{c++} c==2{r=1} END{exit 1-r}' {} \; \
-print
Alternately, if your shell is bash version 4 or above, you can avoid using find and use the bash option globstar:
$ shopt -s globstar
$ awk 'FNR=1{c=0} /#hello1/{c++} /#hello2/{c++} c==2{print FILENAME;nextfile}' **/*
Note: none of this is tested.
If you are not nterested in the number of files also,
then just something along:
find $BASEDIRECTORY -type f -print0 | xargs -0 grep -h PATTERN | wc -l
If you want to count lines containing #hello1 and #hello2 separated by space in a specific file you can:
$ grep -c '#hello1 #hello2' file
If you want to count in more than one file:
$ grep -c '#hello1 #hello2' file1 file2 ...
And if you want to get the gran total:
$ grep -c '#hello1 #hello2' file1 file2 ... | paste -s -d+ - | bc
of course you can let your shell expanding file names. So, for example:
$ grep -c '#hello1 #hello2' *.txt | paste -s -d+ - | bc
or so...
find . -type f | xargs -1 awk '/#hello1/ && /#hello2/{c++} END{print FILENAME, c+0}'
To get X number of files in a directory, I can do:
$ ls -U | head -40000
How would I then delete these 40,000 files? For example, something like:
$ "rm -rf" (ls -U | head -40000)
The tool you need for this is xargs. It will convert standard input into arguments to a command that you specify. Each line of the input is treated as a single argument.
Thus, something like this would work (see the comment below, though, ls shouldn't be parsed this way normally):
ls -U | head -40000 | xargs rm -rf
I would recommend before trying this to start with a small head size and use xargs echo to print out the filenames being passed so you understand what you'll be deleting.
Be aware if you have files with weird characters that this can sometimes be a problem. If you are on a modern GNU system you may also wish to use the arguments to these commands that use null characters to separate each element. Since a filename cannot contain a null character that will safely parse all possible names. I am not aware of a simple way to take the top X items when they are zero separated.
So, for example you can use this to delete all files in a directory
find . -maxdepth 1 -print0 | xargs -0 rm -rf
Use a bash array and slice it. If the number and size of arguments is likely to get close to the system's limits, you can still use xargs to split up the remainder.
files=( * )
printf '%s\0' "${files[#]:0:40000}" | xargs -0 rm
What about using awk as the filter?
find "$FOLDER" -maxdepth 1 -mindepth 1 -print0 \
| awk -v limit=40000 'NR<=limit;NR>limit{exit}' RS="\0" ORS="\0" \
| xargs -0 rm -rf
It will reliably remove at most 40.000 files (or folders). Reliably means regardless of which characters the filenames may contain.
Btw, to get the number of files in a directory reliably you can do:
find FOLDER -mindepth 1 -maxdepth 1 -printf '.' | wc -c
I ended up doing this since my folders were named with sequential numbers. This should also work for alphabetical folders:
ls -r releases/ | sed '1,3d' | xargs -I {} rm -rf releases/{}
Details:
list all the items in the releases/ folder in reverse order
slice off the first 3 items (which would be the newest if numeric/alpha naming)
for each item, rm it
In your case, you can replace ls -r with ls -U and 1,3d with 1,40000d. That should be the same, I believe.
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"
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 :)