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From: frechett@boulder.Colorado.EDU (-=Runaway Daemon=-)
Newsgroups: comp.unix.questions
Subject: Re: getting a field from a line in awk/sed/?
Message-ID: <1991Jan10.085334.19790@csn.org>
Date: 10 Jan 91 08:53:34 GMT
References: <1991Jan9.162817.17038@porthos.cc.bellcore.com> <1991Jan9.225344.19979@informix.com> <5155@idunno.Princeton.EDU>
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In article <5155@idunno.Princeton.EDU> pfalstad@phoenix.Princeton.EDU (Paul John Falstad) writes:
>In article <1991Jan9.225344.19979@informix.com> dberg@informix.com (David I. Berg) writes:
>>Use cut. The syntax is ` cut -fn -d" " ', where n is the number of the
>>field you wish to extract. -d" " specifies spaces as the delimiter between
>>fields. `man cut' will give you all the particulars.
>
>That isn't going to work for his example (output of lpq).  If you do:
>
>$ cut -d' ' -f3 <<EOF
>foo  bar   ble   boz  burn
>EOF
Ah, yes, you are right, but there is a way.  How about if you know that the
first column is only 12 characters wide... Something like 
cut -c1-12 <<EOF
in fact works quite well.  Granted, I would use awk or perl but
cut has its uses too.  Of course there are those without cut but that 
is their problem.

	ian

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