Shell script to count files, then remove oldest files

Question

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

Answer 1

Experiment with this, because I'm not 100% sure it'll work:

cd /backups; ls -at | tail -n +10 | xargs -I{} "rm '{}'"

Answer 2

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

Answer 3

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.

Answer 4

The proper way to do this type of thing is with logrotate.

Answer 5

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 

Answer 6

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