Shell script to count files, then remove oldest files

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

Solution 1:

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.

Solution 2:

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 !