Is there a simple way, in a pretty standard UNIX environment with bash, to run a command to delete all but the most recent X files from a directory?
To give a bit more of a concrete example, imagine some cron job writing out a file (say, a log file or a tar-ed up backup) to a directory every hour. I'd like a way to have another cron job running which would remove the oldest files in that directory until there are less than, say, 5.
And just to be clear, there's only one file present, it should never be deleted.
The problems with the existing answers:
inability to handle filenames with embedded spaces or newlines.
in the case of solutions that invoke rm directly on an unquoted command substitution (rm `...`), there's an added risk of unintended globbing.
inability to distinguish between files and directories (i.e., if directories happened to be among the 5 most recently modified filesystem items, you'd effectively retain fewer than 5 files, and applying rm to directories will fail).
wnoise's answer addresses these issues, but the solution is GNU-specific (and quite complex).
Here's a pragmatic, POSIX-compliant solution that comes with only one caveat: it cannot handle filenames with embedded newlines - but I don't consider that a real-world concern for most people.
For the record, here's the explanation for why it's generally not a good idea to parse ls output: http://mywiki.wooledge.org/ParsingLs
ls -tp | grep -v '/$' | tail -n +6 | xargs -I {} rm -- {}
Note: This command operates in the current directory; to target a directory explicitly, use a subshell ((...)) with cd:
(cd /path/to && ls -tp | grep -v '/$' | tail -n +6 | xargs -I {} rm -- {})
The same applies analogously to the commands below.
The above is inefficient, because xargs has to invoke rm separately for each filename.
However, your platform's specific xargs implementation may allow you to solve this problem:
A solution that works with GNU xargs is to use -d '\n', which makes xargs consider each input line a separate argument, yet passes as many arguments as will fit on a command line at once:
ls -tp | grep -v '/$' | tail -n +6 | xargs -d '\n' -r rm --
Note: Option -r (--no-run-if-empty) ensures that rm is not invoked if there's no input.
A solution that works with both GNU xargs and BSD xargs (including on macOS) - though technically still not POSIX-compliant - is to use -0 to handle NUL-separated input, after first translating newlines to NUL (0x0) chars., which also passes (typically) all filenames at once:
ls -tp | grep -v '/$' | tail -n +6 | tr '\n' '\0' | xargs -0 rm --
Explanation:
ls -tp prints the names of filesystem items sorted by how recently they were modified , in descending order (most recently modified items first) (-t), with directories printed with a trailing / to mark them as such (-p).
Note: It is the fact that ls -tp always outputs file / directory names only, not full paths, that necessitates the subshell approach mentioned above for targeting a directory other than the current one ((cd /path/to && ls -tp ...)).
grep -v '/$' then weeds out directories from the resulting listing, by omitting (-v) lines that have a trailing / (/$).
Caveat: Since a symlink that points to a directory is technically not itself a directory, such symlinks will not be excluded.
tail -n +6 skips the first 5 entries in the listing, in effect returning all but the 5 most recently modified files, if any.
Note that in order to exclude N files, N+1 must be passed to tail -n +.
xargs -I {} rm -- {} (and its variations) then invokes on rm on all these files; if there are no matches at all, xargs won't do anything.
xargs -I {} rm -- {} defines placeholder {} that represents each input line as a whole, so rm is then invoked once for each input line, but with filenames with embedded spaces handled correctly.
-- in all cases ensures that any filenames that happen to start with - aren't mistaken for options by rm.
A variation on the original problem, in case the matching files need to be processed individually or collected in a shell array:
# One by one, in a shell loop (POSIX-compliant):
ls -tp | grep -v '/$' | tail -n +6 | while IFS= read -r f; do echo "$f"; done
# One by one, but using a Bash process substitution (<(...),
# so that the variables inside the `while` loop remain in scope:
while IFS= read -r f; do echo "$f"; done < <(ls -tp | grep -v '/$' | tail -n +6)
# Collecting the matches in a Bash *array*:
IFS=$'\n' read -d '' -ra files < <(ls -tp | grep -v '/$' | tail -n +6)
printf '%s\n' "${files[#]}" # print array elements
Remove all but 5 (or whatever number) of the most recent files in a directory.
rm `ls -t | awk 'NR>5'`
(ls -t|head -n 5;ls)|sort|uniq -u|xargs rm
This version supports names with spaces:
(ls -t|head -n 5;ls)|sort|uniq -u|sed -e 's,.*,"&",g'|xargs rm
Simpler variant of thelsdj's answer:
ls -tr | head -n -5 | xargs --no-run-if-empty rm
ls -tr displays all the files, oldest first (-t newest first, -r reverse).
head -n -5 displays all but the 5 last lines (ie the 5 newest files).
xargs rm calls rm for each selected file.
find . -maxdepth 1 -type f -printf '%T# %p\0' | sort -r -z -n | awk 'BEGIN { RS="\0"; ORS="\0"; FS="" } NR > 5 { sub("^[0-9]*(.[0-9]*)? ", ""); print }' | xargs -0 rm -f
Requires GNU find for -printf, and GNU sort for -z, and GNU awk for "\0", and GNU xargs for -0, but handles files with embedded newlines or spaces.
All these answers fail when there are directories in the current directory. Here's something that works:
find . -maxdepth 1 -type f | xargs -x ls -t | awk 'NR>5' | xargs -L1 rm
This:
works when there are directories in the current directory
tries to remove each file even if the previous one couldn't be removed (due to permissions, etc.)
fails safe when the number of files in the current directory is excessive and xargs would normally screw you over (the -x)
doesn't cater for spaces in filenames (perhaps you're using the wrong OS?)
ls -tQ | tail -n+4 | xargs rm
List filenames by modification time, quoting each filename. Exclude first 3 (3 most recent). Remove remaining.
EDIT after helpful comment from mklement0 (thanks!): corrected -n+3 argument, and note this will not work as expected if filenames contain newlines and/or the directory contains subdirectories.
Ignoring newlines is ignoring security and good coding. wnoise had the only good answer. Here is a variation on his that puts the filenames in an array $x
while IFS= read -rd ''; do
x+=("${REPLY#* }");
done < <(find . -maxdepth 1 -printf '%T# %p\0' | sort -r -z -n )
For Linux (GNU tools), an efficient & robust way to keep the n newest files in the current directory while removing the rest:
n=5
find . -maxdepth 1 -type f -printf '%T# %p\0' |
sort -z -nrt ' ' -k1,1 |
sed -z -e "1,${n}d" -e 's/[^ ]* //' |
xargs -0r rm -f
For BSD, find doesn't have the -printf predicate, stat can't output NULL bytes, and sed + awk can't handle NULL-delimited records.
Here's a solution that doesn't support newlines in paths but that safeguards against them by filtering them out:
#!/bin/bash
n=5
find . -maxdepth 1 -type f ! -path $'*\n*' -exec stat -f '%.9Fm %N' {} + |
sort -nrt ' ' -k1,1 |
awk -v n="$n" -F'^[^ ]* ' 'NR > n {printf "%s%c", $2, 0}' |
xargs -0 rm -f
note: I'm using bash because of the $'\n' notation. For sh you can define a variable containing a literal newline and use it instead.
Solution for UNIX & Linux (inspired from AIX/HP-UX/SunOS/BSD/Linux ls -b):
Some platforms don't provide find -printf, nor stat, nor support NUL-delimited records with stat/sort/awk/sed/xargs. That's why using perl is probably the most portable way to tackle the problem, because it is available by default in almost every OS.
I could have written the whole thing in perl but I didn't. I only use it for substituting stat and for encoding-decoding-escaping the filenames. The core logic is the same as the previous solutions and is implemented with POSIX tools.
note: perl's default stat has a resolution of a second, but starting from perl-5.8.9 you can get sub-second resolution with the stat function of the module Time::HiRes (when both the OS and the filesystem support it). That's what I'm using here; if your perl doesn't provide it then you can remove the ‑MTime::HiRes=stat from the command line.
n=5
find . '(' -name '.' -o -prune ')' -type f -exec \
perl -MTime::HiRes=stat -le '
foreach (#ARGV) {
#st = stat($_);
if ( #st > 0 ) {
s/([\\\n])/sprintf( "\\%03o", ord($1) )/ge;
print sprintf( "%.9f %s", $st[9], $_ );
}
else { print STDERR "stat: $_: $!"; }
}
' {} + |
sort -nrt ' ' -k1,1 |
sed -e "1,${n}d" -e 's/[^ ]* //' |
perl -l -ne '
s/\\([0-7]{3})/chr(oct($1))/ge;
s/(["\n])/"\\$1"/g;
print "\"$_\"";
' |
xargs -E '' sh -c '[ "$#" -gt 0 ] && rm -f "$#"' sh
Explanations:
For each file found, the first perl gets the modification time and outputs it along the encoded filename (each newline and backslash characters are replaced with the literals \012 and \134 respectively).
Now each time filename is guaranteed to be single-line, so POSIX sort and sed can safely work with this stream.
The second perl decodes the filenames and escapes them for POSIX xargs.
Lastly, xargs calls rm for deleting the files. The sh command is a trick that prevents xargs from running rm when there's no files to delete.
I realize this is an old thread, but maybe someone will benefit from this. This command will find files in the current directory :
for F in $(find . -maxdepth 1 -type f -name "*_srv_logs_*.tar.gz" -printf '%T# %p\n' | sort -r -z -n | tail -n+5 | awk '{ print $2; }'); do rm $F; done
This is a little more robust than some of the previous answers as it allows to limit your search domain to files matching expressions. First, find files matching whatever conditions you want. Print those files with the timestamps next to them.
find . -maxdepth 1 -type f -name "*_srv_logs_*.tar.gz" -printf '%T# %p\n'
Next, sort them by the timestamps:
sort -r -z -n
Then, knock off the 4 most recent files from the list:
tail -n+5
Grab the 2nd column (the filename, not the timestamp):
awk '{ print $2; }'
And then wrap that whole thing up into a for statement:
for F in $(); do rm $F; done
This may be a more verbose command, but I had much better luck being able to target conditional files and execute more complex commands against them.
If the filenames don't have spaces, this will work:
ls -C1 -t| awk 'NR>5'|xargs rm
If the filenames do have spaces, something like
ls -C1 -t | awk 'NR>5' | sed -e "s/^/rm '/" -e "s/$/'/" | sh
Basic logic:
get a listing of the files in time order, one column
get all but the first 5 (n=5 for this example)
first version: send those to rm
second version: gen a script that will remove them properly
With zsh
Assuming you don't care about present directories and you will not have more than 999 files (choose a bigger number if you want, or create a while loop).
[ 6 -le `ls *(.)|wc -l` ] && rm *(.om[6,999])
In *(.om[6,999]), the . means files, the o means sort order up, the m means by date of modification (put a for access time or c for inode change), the [6,999] chooses a range of file, so doesn't rm the 5 first.
Adaptation of #mklement0's excellent answer with some parameters and without needing to navigate to the folder containing the files to be deleted...
TARGET_FOLDER="/my/folder/path"
FILES_KEEP=5
ls -tp "$TARGET_FOLDER"**/* | grep -v '/$' | tail -n +$((FILES_KEEP+1)) | xargs -d '\n' -r rm --
[Ref(s).: https://stackoverflow.com/a/3572628/3223785 ]
Thanks! 😉
found interesting cmd in Sed-Onliners - Delete last 3 lines - fnd it perfect for another way to skin the cat (okay not) but idea:
#!/bin/bash
# sed cmd chng #2 to value file wish to retain
cd /opt/depot
ls -1 MyMintFiles*.zip > BigList
sed -n -e :a -e '1,2!{P;N;D;};N;ba' BigList > DeList
for i in `cat DeList`
do
echo "Deleted $i"
rm -f $i
#echo "File(s) gonzo "
#read junk
done
exit 0
Removes all but the 10 latest (most recents) files
ls -t1 | head -n $(echo $(ls -1 | wc -l) - 10 | bc) | xargs rm
If less than 10 files no file is removed and you will have :
error head: illegal line count -- 0
To count files with bash
I needed an elegant solution for the busybox (router), all xargs or array solutions were useless to me - no such command available there. find and mtime is not the proper answer as we are talking about 10 items and not necessarily 10 days. Espo's answer was the shortest and cleanest and likely the most unversal one.
Error with spaces and when no files are to be deleted are both simply solved the standard way:
rm "$(ls -td *.tar | awk 'NR>7')" 2>&-
Bit more educational version: We can do it all if we use awk differently. Normally, I use this method to pass (return) variables from the awk to the sh. As we read all the time that can not be done, I beg to differ: here is the method.
Example for .tar files with no problem regarding the spaces in the filename. To test, replace "rm" with the "ls".
eval $(ls -td *.tar | awk 'NR>7 { print "rm \"" $0 "\""}')
Explanation:
ls -td *.tar lists all .tar files sorted by the time. To apply to all the files in the current folder, remove the "d *.tar" part
awk 'NR>7... skips the first 7 lines
print "rm \"" $0 "\"" constructs a line: rm "file name"
eval executes it
Since we are using rm, I would not use the above command in a script! Wiser usage is:
(cd /FolderToDeleteWithin && eval $(ls -td *.tar | awk 'NR>7 { print "rm \"" $0 "\""}'))
In the case of using ls -t command will not do any harm on such silly examples as: touch 'foo " bar' and touch 'hello * world'. Not that we ever create files with such names in real life!
Sidenote. If we wanted to pass a variable to the sh this way, we would simply modify the print (simple form, no spaces tolerated):
print "VarName="$1
to set the variable VarName to the value of $1. Multiple variables can be created in one go. This VarName becomes a normal sh variable and can be normally used in a script or shell afterwards. So, to create variables with awk and give them back to the shell:
eval $(ls -td *.tar | awk 'NR>7 { print "VarName=\""$1"\"" }'); echo "$VarName"
leaveCount=5
fileCount=$(ls -1 *.log | wc -l)
tailCount=$((fileCount - leaveCount))
# avoid negative tail argument
[[ $tailCount < 0 ]] && tailCount=0
ls -t *.log | tail -$tailCount | xargs rm -f
I made this into a bash shell script. Usage: keep NUM DIR where NUM is the number of files to keep and DIR is the directory to scrub.
#!/bin/bash
# Keep last N files by date.
# Usage: keep NUMBER DIRECTORY
echo ""
if [ $# -lt 2 ]; then
echo "Usage: $0 NUMFILES DIR"
echo "Keep last N newest files."
exit 1
fi
if [ ! -e $2 ]; then
echo "ERROR: directory '$1' does not exist"
exit 1
fi
if [ ! -d $2 ]; then
echo "ERROR: '$1' is not a directory"
exit 1
fi
pushd $2 > /dev/null
ls -tp | grep -v '/' | tail -n +"$1" | xargs -I {} rm -- {}
popd > /dev/null
echo "Done. Kept $1 most recent files in $2."
ls $2|wc -l
Modified version of the answer of #Fabien if you want to specify a path. Useful if you're running the script elsewhere.
ls -tr /path/foo/ | head -n -5 | xargs -I% --no-run-if-empty rm /path/foo/%
I've got a command to perform a series of commands that produce a variable output string such as 123456. I want to pipe that to a sed command replacing a known string in a csv file that looks like this:
Fred,Wilma,Betty,Barney
However, the command below does not work and I haven't found any other references to using pipe values as the variable for a replace.
How does this code change if the values in the csv are in a random order and I always want to change the second value?
Example code:
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 > test.txt |
sed -i -e '/Wilma/ r test.txt' -e 's/Wilma//' input.csv
Contents of input.csv should become: Fred,0d522cd316,Betty,Barney
Okay, in
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 > test.txt | sed -i -e '/Wilma/ r test.txt' -e 's/Wilma//' input.csv
you have a bug. That "> test.txt" after cut is going to eat your stdin on sed, so things go weird with that pipe afterwards taking stdin. You don't want a pipe there, or you don't want to redirect to a file.
The way to take piped stdin and use it as a parameter in a command is through xargs.
find / -iname awk 2>/dev/null | sha256sum | cut -c1-10 | xargs --replace=INSERTED -- sed -i -e 's/Wilma/INSERTED/' input.csv
(...though that find|shasum is suspect too, in that the order of files is random(ish) and it matters for a reliable sum. You prpobably mean to "|sort" after find.)
(Some would sed -i -e "s/Wilma/$(find|sort|shasum|cut)" f, but I ain't among them. Animals.)
For replacing a fixed string like "Wilma", try:
sed -i 's/Wilma/'"$(find / -iname awk 2>/dev/null |
sha256sum | cut -c1-10)"'/' input.csv
To replace the 2nd field no matter what's in it, try:
sed -i 's/[^,]*/'"$(find / -iname awk 2>/dev/null |
sha256sum | cut -c1-10)"'/2' input.csv
I have a directory with thousands of files (100K for now). When I use wc -l ./*, I'll get:
c1 ./test1.txt
c2 ./test2.txt
...
cn ./testn.txt
c1+c2+...+cn total
Because there are a lot of files in the directory, I just want to see the total count and not the details. Is there any way to do so?
I tried several ways and I got following error:
Argument list too long
If what you want is the total number of lines and nothing else, then I would suggest the following command:
cat * | wc -l
This catenates the contents of all of the files in the current working directory and pipes the resulting blob of text through wc -l.
I find this to be quite elegant. Note that the command produces no extraneous output.
UPDATE:
I didn't realize your directory contained so many files. In light of this information, you should try this command:
for file in *; do cat "$file"; done | wc -l
Most people don't know that you can pipe the output of a for loop directly into another command.
Beware that this could be very slow. If you have 100,000 or so files, my guess would be around 10 minutes. This is a wild guess because it depends on several parameters that I'm not able to check.
If you need something faster, you should write your own utility in C. You could make it surprisingly fast if you use pthreads.
Hope that helps.
LAST NOTE:
If you're interested in building a custom utility, I could help you code one up. It would be a good exercise, and others might find it useful.
Credit: this builds on #lifecrisis's answer, and extends it to handle large numbers of files:
find . -maxdepth 1 -type f -exec cat {} + | wc -l
find will find all of the files in the current directory, break them into groups as large as can be passed as arguments, and run cat on the groups.
awk 'END {print NR" total"}' ./*
Would be an interesting comparison to find out how many lines don't end with a new line.
Combining the awk and Gordon’s find solutions and avoiding the "." files.
find ./* -maxdepth 0 -type f -exec awk 'END {print NR}' {} +
No idea if this is better or worse but it does give a more accurate count (for me) and does not count lines in "." files. Using ./* is just a guess that appears to work.
Still need depth and ./* requires "0" depth.
I did get the same result with the "cat" and "awk" solutions (using the same find) since the "cat *" takes care of the new line issue. I don't have a directory with enough files to measure time. Interesting, I'm liking the "cat" solution.
This will give you the total count for all the files (including hidden files) in your current directory :
$ find . -maxdepth 1 -type f | xargs wc -l | grep total
1052 total
To count for files excluding hidden files use :
find . -maxdepth 1 -type f -not -path "*/\.*" | xargs wc -l | grep total
(Apologies for adding this as an answer—but I do not have enough reputation for commenting.)
A comment on #lifecrisis's answer. Perhaps cat is slowing things down a bit. We could replace cat by wc -l and then use awkto add the numbers. (This could be faster since much less data needs to go throught the pipe.)
That is
for file in *; do wc -l "$file"; done | awk '{sum += $1} END {print sum}'
instead of
for file in *; do cat "$file"; done | wc -l
(Disclaimer: I am not incorporating many of the improvements in other answers, but I thought the point was valid enough to write down.)
Here are my results for comparison (I ran the newer version first so that any cache effects would go against the newer candidate).
$ time for f in `seq 1 1500`; do head -c 5M </dev/urandom >myfile-$f |sed -e 's/\(................\)/\1\n/g'; done
real 0m50.360s
user 0m4.040s
sys 0m49.489s
$ time for file in myfile-*; do wc -l "$file"; done | awk '{sum += $1} END {print sum}'
30714902
real 0m3.455s
user 0m2.093s
sys 0m1.515s
$ time for file in myfile-*; do cat "$file"; done | wc -l
30714902
real 0m4.481s
user 0m2.544s
sys 0m4.312s
iF you want to know only total number Lines in directory excluding total line
ls -ltr | sed -n '/total/!p' | awk '{print NR}'
Previous comment will give total count of lines which includes only count of lines in all files
Below command will provide the total count of lines from all files in path
for i in `ls- ltr | awk ‘$1~”^-rw”{print $9}’`; do wc -l $I | awk ‘{print $1}’; done >>/var/tmp/filelinescount.txt
Cat /var/tmp/filelinescount.txt| sed -r “s/\s+//g”|tr “\n” “+”| sed “s:+$::g”| sed ’s/^/“/g’| sed ’s/$/“/g’ | awk ‘{print “echo” “ “ $0”+bc”}’| sh
I've got the following line:
wc -l ./*.txt | sort -rn
i want to cut the file extension. So with this code i've got the output:
number filename.txt
for all my .txt-files in the .-directory. But I want the output without the file-extension, like this:
number filename
I tried a pipe with cut for different kinds of parameter, but all i got was to cut the whole filename with this command.
wc -l ./*.txt | sort -rn | cut -f 1 -d '.'
Assuming you don't have newlines in your filename you can use sed to strip out ending .txt:
wc -l ./*.txt | sort -rn | sed 's/\.txt$//'
unfortunately, cut doesn't have a syntax for extracting columns according to an index from the end. One (somewhat clunky) trick is to use rev to reverse the line, apply cut to it and then rev it back:
wc -l ./*.txt | sort -rn | rev | cut -d'.' -f2- | rev
Using sed in more generic way to cut off whatever extension the files have:
$ wc -l *.txt | sort -rn | sed 's/\.[^\.]*$//'
14 total
8 woc
3 456_base
3 123_base
0 empty_base
A better approach using proper mime type (what is the extension of tar.gz or such multi extensions ? )
#!/bin/bash
for file; do
case $(file -b $file) in
*ASCII*) echo "this is ascii" ;;
*PDF*) echo "this is pdf" ;;
*) echo "other cases" ;;
esac
done
This is a POC, not tested, feel free to adapt/improve/modify
I need help using xargs(1) and bc(1) in the same line. I can do it multiple lines, but I really want to find a solution in one line.
Here is the problem: The following line will print the size of a file.txt
ls -l file.txt | cut -d" " -f5
And, the following line will print 1450 (which is obviously 1500 - 50)
echo '1500-50' | bc
Trying to add those two together, I do this:
ls -l file.txt | cut -d" " -f5 | xargs -0 -I {} echo '{}-50' | bc
The problem is, it's not working! :)
I know that xargs is probably not the right command to use, but it's the only command I can find who can let me decide where to put the argument I get from the pipe.
This is not the first time I'm having issues with this kind of problem. It will be much of a help..
Thanks
If you do
ls -l file.txt | cut -d" " -f5 | xargs -0 -I {} echo '{}-50'
you will see this output:
23
-50
This means, that bc does not see a complete expression.
Just use -n 1 instead of -0:
ls -l file.txt | cut -d" " -f5 | xargs -n 1 -I {} echo '{}-50'
and you get
23-50
which bc will process happily:
ls -l file.txt | cut -d" " -f5 | xargs -n 1 -I {} echo '{}-50' | bc
-27
So your basic problem is, that -0 expects not lines but \0 terminated strings. And hence the newline(s) of the previous commands in the pipe garble the expression of bc.
This might work for you:
ls -l file.txt | cut -d" " -f5 | sed 's/.*/&-50/' | bc
Infact you could remove the cut:
ls -l file.txt | sed -r 's/^(\S+\s+){4}(\S+).*/\2-50/' | bc
Or use awk:
ls -l file.txt | awk '{print $5-50}'
Parsing output from the ls command is not the best idea. (really).
you can use many other solutions, like:
find . -name file.txt -printf "%s\n"
or
stat -c %s file.txt
or
wc -c <file.txt
and can use bash arithmetics, for avoid unnecessary slow process forks, like:
find . -type f -print0 | while IFS= read -r -d '' name
do
size=$(wc -c <$name)
s50=$(( $size - 50 ))
echo "the file=$name= size:$size minus 50 is: $s50"
done
Here is another solution, which only use one external command: stat:
file_size=$(stat -c "%s" file.txt) # Get the file size
let file_size=file_size-50 # Subtract 50
If you really want to combine them into one line:
let file_size=$(stat -c "%s" file.txt)-50
The stat command gets you the file size in bytes. The syntax above is for Linux (I tested against Ubuntu). On the Mac the syntax is a little different:
let file_size=$(stat -f "%z" mini.csv)-50