Remove parts of a path using parameter expansion… but how?

Question

I need to remove the part

man1/bmtoa.1.gz

from the full path

/usr/share/man/man1/bmtoa.1.gz

so that it results in

/usr/share/man/

I've tried ${path#[!/]*/*} and ${path#/*/*} but both result in wrong paths.

Answer 1

this will work ${path%/*/*}, but works from the back to the front. here is a good how-to for chopping strings in bash.

Answer 2

If you could deal with /usr/share/man/man1, then you could do dirname /usr/share/man/man1/bmtoa.1.gz, but since you stated that you need the man1 to also come off, so you'll have to daisy-chain them:

dirname `dirname /usr/share/man1/bmtoa.1.gz`