Aiza <aiz...@comclark.com> writes: > I have a file containing this > > drwxrwxr-x 14 89987 546 512 Jun 6 2009 7.2-RELEASE > drwxrwxr-x 14 89987 546 512 Mar 23 04:59 7.3-RELEASE > drwxrwxr-x 13 89987 546 512 Nov 23 2009 8.0-RELEASE > drwxrwxr-x 13 89987 546 512 Jul 1 04:56 8.1-RC2 > > I want to strip off everything to the left of the release > version so I end up with this. > > 7.2-RELEASE > 7.3-RELEASE > 8.0-RELEASE > 8.1-RC2 > > How would I code to do this?
Use... - glob expansion + echo builtin, e.g. $ cd /path/to/blah && echo * or $ cd /path/to/blah && for f in *; do echo $f; done - field splitting, e.g. $ ls -l | while read $(while [ $((i+=1)) -le 9 ]; do echo p$i; done); do echo $p9; done - stat(1) if you need not only filename but e.g. date Of course you can use smth like cut/sed/awk/whatever but they'll only make your script slower if you use them often. _______________________________________________ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to "freebsd-questions-unsubscr...@freebsd.org"