Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!usc!wuarchive!psuvax1!psuvm!blekul11!ffaac09
From: FFAAC09@cc1.kuleuven.ac.be (Nicole Delbecque & Paul Bijnens)
Newsgroups: comp.editors
Subject: Re: Regular Expression (vi type) question...
Message-ID: <91088.120845FFAAC09@cc1.kuleuven.ac.be>
Date: 29 Mar 91 12:07:45 GMT
References: <1991Mar28.191204.12295@cbnewsl.att.com>
Distribution: na
Organization: K.U.Leuven - Academic Computing Center
Lines: 36

In article <1991Mar28.191204.12295@cbnewsl.att.com>, sgilley@cbnewsl.att.com
(The Idealistic Cynic) says:
>
>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.)

No need to assume there is a dot:

    :%s/\.[^.]*$//

in english:
    on all lines, substitute a literal dot, followed by anything
    but a dot, only at the end of a line, by nothing.
The substitution happens only on the lines that match.
Works with me (SYSV3.2 vi version 3.9 2/9/83).
--
Polleke   (Paul Bijnens)
Linguistics dept., K. University Leuven, Belgium
FFAAC09@cc1.kuleuven.ac.be