[LUG] grep/sed print matching group only

[Simon Williams <simon@xxxxxxxxxxxxxxxxxxx>]

Afternoon all.

How can I make grep or sed print a matching group *only*?
So far the only solution I have been able to find is to use grep to pick 
out the line and then use sed to substitute the entire line for just the 
matching group, like this:

grep '^foo .* bar$' | sed 's/foo \(.*\) bar/\1/'

This is clearly stupid and there must be a better way. Why can't I do 
something like this?

grep '^foo (.*) bar$' -[some option]
or:
sed '/^foo (.*) bar$/\1/'

This isn't the first time I've been driven completely insane by this 
problem. I don't understand why it's so difficult.

Can anyone help me?

Thanks.
Simon

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