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.

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score 1 · Answer 1 · answered Jan 20 '10 at 12:59

1

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

answered Jan 20 '10 at 12:59

Christian

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score 1 · Answer 2 · answered Jan 20 '10 at 13:20

1

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`

answered Jan 20 '10 at 13:20

Kevin M

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It doesn't use shell expansion as requested by to me dirname looks much saner and simpler. – Jason Tan Apr 25 '10 at 15:18

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

2 Answers2