Can grep output only specified groupings that match?

GNU grep has the -P option for perl-style regexes, and the -o option to print only what matches the pattern. These can be combined using look-around assertions (described under Extended Patterns in the perlre manpage) to remove part of the grep pattern from what is determined to have matched for the purposes of -o.

$ grep -oP 'foobar \K\w+' test.txt
bash
happy
$

The \K is the short-form (and more efficient form) of (?<=pattern) which you use as a zero-width look-behind assertion before the text you want to output. (?=pattern) can be used as a zero-width look-ahead assertion after the text you want to output.

For instance, if you wanted to match the word between foo and bar, you could use:

$ grep -oP 'foo \K\w+(?= bar)' test.txt

or (for symmetry)

$ grep -oP '(?<=foo )\w+(?= bar)' test.txt

    sed -n "s/^.*foobar\s*\(\S*\).*$/\1/p"

-n     suppress printing
s      substitute
^.*    anything before foobar
foobar initial search match
\s*    any white space character (space)
\(     start capture group
\S*    capture any non-white space character (word)
\)     end capture group
.*$    anything after the capture group
\1     substitute everything with the 1st capture group
p      print it

Standard grep can't do this, but recent versions of GNU grep can. You can turn to sed, awk or perl. Here are a few examples that do what you want on your sample input; they behave slightly differently in corner cases.

Replace foobar word other stuff by word, print only if a replacement is done.

sed -n -e 's/^foobar \([[:alnum:]]\+\).*/\1/p'

If the first word is foobar, print the second word.

awk '$1 == "foobar" {print $2}'

Strip foobar if it's the first word, and skip the line otherwise; then strip everything after the first whitespace and print.

perl -lne 's/^foobar\s+// or next; s/\s.*//; print'