Search and remove with regular expression

Question

How do you use regular expression in bash to search any given string and remove occurrence of particular word? This is the expression I've come up with

^([A-Za-z-]+)(-).*(el6.*)$

What I would like to accomplish is, in a given string like cjkuni-ukai-fonts-0.2.20080216.1-35.el6.noarch, the expression should remove cjkuni-ukai-fonts- and el6.noarch, and leave only 0.2.20080216.1-35. As I'm very limited in what I can install on the system, I cannot use perl or other non basic commands. Any help would be appreciated.

P.S : I've tried sed, but I couldn't get it work.

Answer 1

If you really must do it in pure bash:

$ foo="cjkuni-ukai-fonts-0.2.20080216.1-35.el6.noarch"
$ [[ $foo =~ [0-9.]+-[0-9]* ]] && echo $BASH_REMATCH
0.2.20080216.1-35

If you're OK with a sed solution:

$ sed 's/.*-\([0-9.]*-[0-9]*\).*/\1/' <<<$foo
0.2.20080216.1-35

If you have access to GNU grep, you could also do:

$ grep -oE -- '[0-9.]+-[0-9]*' <<<$foo
0.2.20080216.1-35

or

$ grep -oP -- '[.\d]+-\d+' <<<$foo
0.2.20080216.1-35

Finally, since you're running a CentOS system, it is a fairly safe bet that Perl is installed, so you could also do:

$ perl -pe 's/.*?([0-9.]+-[0-9]*).*/$1/' <<<$foo
0.2.20080216.1-35