Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!tut.cis.ohio-state.edu!pt.cs.cmu.edu!andrew.cmu.edu!jaap+
From: jaap+@andrew.cmu.edu (Jaap Akkerhuis)
Newsgroups: comp.unix.questions
Subject: Re: extracting "rest of line" in AWK
Message-ID: <YYxMUtK00jbYEFS2cc@andrew.cmu.edu>
Date: 25 Aug 89 17:56:09 GMT
References: <3368@blake.acs.washington.edu>
Distribution: na
Organization: Information Technology Center, Carnegie Mellon, Pittsburgh, PA
Lines: 28
In-Reply-To: <3368@blake.acs.washington.edu>

> Excerpts from netnews.comp.unix.questions: 25-Aug-89 extracting "rest of
> line" i.. Mark Bader@cac.washingto (670)

> Does anyone know of a way to extract the "rest of the line" in awk..
> e.g. I have a line that looks like

> %% Heading "This is a Graph Heading"

> and I want to do a {print $3} to get the "This is a Graph Heading" part,
> but this obviously dosen't work.   Is there a way to do this?  

> If there were a shift command, then I could shift off the first 2 fields,
but is there a way to simulate this?

Allthough it is a bit ugly, you can empty the first two fields, so
something like:

	{ $1="\0"; $2="\0"; print $0 }

Of course, now you will have two spaces in front of the line printed. If
you want to get rid of this you need a for loop in the style of "{ for(i
= 3; i < NF; i++) printf("%s ", $i); print ""}" or just filter them out
with an "awk '{ $1=""; $2=""; print $0 }' | sed 's/^  //'".

There are tons of other ways to do this as well.


	jaap