sed - How to get "all but the last N fields/things" in shell script? -

i encounter case need beginning of string without last part.

for example, host part of fqdn without domain name:

www.example.org -> www foo.bar.example.org -> foo.bar foo.bar.baz.some.domain.name -> foo.bar.baz.some

(the same applies other examples need strip end of input.)

i use mix of cut , rev this, such as:

echo "foo.bar.example.org" | rev | cut -d "." -f 3- | rev

this trick, wonder if there better way this?

with awk do:

$ awk '{nf=nf-2}1' fs=. ofs=. file www foo.bar foo.bar.baz.some

generalize last n feilds n=2:

$ awk '{nf=nf-n}1' fs=. ofs=. n=2 file www foo.bar foo.bar.baz.some

with sed do:

$ sed 's/[.][^.]*[.][^.]*$//' file www foo.bar foo.bar.baz.some

generalized:

$ sed -r 's/([.][^.]*){2}$//' file www foo.bar foo.bar.baz.some

Roma