[bash] Delete all but the most recent X files in bash

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.

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

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.

Remove all but 5 (or whatever number) of the most recent files in a directory.

rm `ls -t | awk 'NR>5'`

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.

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

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.

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.

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

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:

1. works when there are directories in the current directory

2. tries to remove each file even if the previous one couldn't be removed (due to permissions, etc.)

3. fails safe when the number of files in the current directory is excessive and xargs would normally screw you over (the -x)

4. doesn't cater for spaces in filenames (perhaps you're using the wrong OS?)

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"

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.

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

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 )

(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