Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!csd4.csd.uwm.edu!uakari.primate.wisc.edu!xanth!mcnc!decvax!ima!cfisun!lakart!dg
From: dg@lakart.UUCP (David Goodenough)
Newsgroups: comp.unix.questions
Subject: Re: extracting "rest of line" in AWK
Message-ID: <671@lakart.UUCP>
Date: 27 Aug 89 13:28:34 GMT
References: <3368@blake.acs.washington.edu>
Distribution: na
Organization: Lakart Corporation, Newton, MA
Lines: 35

mbader@cac.washington.edu (Mark Bader) asks:
> 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?  

It's fairly grotesque, a lot of work, but the following should work,
assuming you're not to worried about spaces.

	{ for (i = 3; i <= NF; i++) {
			if (i == NF)
			 {
			    a = "\n";
			 }
			else
			 {
			    a = " ";
			 }
			printf "%s%s", $i, a; } }

Or somesuch. I'm making the above up as I type here, so I can't guarantee it,
but it should provide a starting point. An alternative approah is to use
length($1) and length($2) and substr(), but again that assumes a lot about
the input record.

Enough rambling, I'm not really an awk guru. We now return you to your
regular news articles.
-- 
	dg@lakart.UUCP - David Goodenough		+---+
						IHS	| +-+-+
	....... !harvard!xait!lakart!dg			+-+-+ |
AKA:	dg%lakart.uucp@xait.xerox.com			  +---+