Bash one-liner to delete only old kernels

I've seen lots of threads on how to free space on the /boot partition and that is my objective as well. However, I'm only interested in deleting old kernels and not each one of them but the current one.

I need the solution to be a one-liner since I'll be running the script from Puppet and I don't want to have extra files lying around. So far I got the following:

dpkg -l linux-* | awk '/^ii/{print $2}' | egrep [0-9] | sort -t- -k3,4 --version-sort -r | sed -e "1,/$(uname -r | cut -f1,2 -d"-")/d" | grep -v -e `uname -r | cut -f1,2 -d"-"` | xargs sudo apt-get -y purge

To be more precise, what it does at the moment is the following:

• List all the linux-* packages and print their names.


• Only list the ones that have numbers and sort them, returning the reverse result. This way, older kernels are listed last.


• Print only the results that go after the current kernel


• Since there are some linux-{image,headers} results, make sure I won't purge anything related to my current kernel



• Call apt to purge

This works, but I'm sure the solution can be more elegant and that it's safe for a production environment, since at least 20 of our servers run Ubuntu.

Thanks for your time,
Alejandro.