Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!att!cbnewsl!sgilley
From: sgilley@cbnewsl.att.com (The Idealistic Cynic)
Newsgroups: comp.editors
Subject: Regular Expression (vi type) question...
Message-ID: <1991Mar28.191204.12295@cbnewsl.att.com>
Date: 28 Mar 91 19:12:04 GMT
Distribution: na
Organization: AT&T, Piscataway, New Jersey
Lines: 29

Can a single regular expression be used to delete
all characters including and after the last "."
in a line?

Input might be:

test
stay.c
junk.1.c

And I need the output to be:

test
stay
junk.1

I've played with this a while, and can't seem to come
up with an answer that will work. (It's easy if you
assume all lines will have at least one "." in them..
what I can't get is if they don't.)

Thanks,

Sean.

---
Sean L. Gilley
attmail!mycroft!slg  attmail!sgilley
201 805 9088 (h)    201 457 5403 (w)