From kevin1a at varlog.net  Mon Apr 26 22:16:04 2004
From: kevin1a at varlog.net (Kevin Brouelette)
Date: Mon, 26 Apr 2004 22:16:04 -0700
Subject: [OCLUG-devel] Learning awk
Message-ID: <1083042964.1490.20.camel@athlon.kblan.com>


Hello,

I'm spending my weeknights this week brushing up on bash shell.
The book I'm working with is Addison-Wesley's 'Linux & Unix shell 
Programing' by David Tansley.

Chapter 9 is about 'awk'. 
After reading the chapter,
I thought I would try to use awk to grab the users 
which their UID from /etc/passwd is greater than 999.
[Slackware UIDs start at 1000]

So I try this:
awk -F: '{if($2 > 999) print$0"\t"$2}' /etc/passwd

hoping to print the name, a 'tab', and the UID.

It echo's the whole /etc/passwd file and no errors.
Am I close or am I off base with trying to do this with awk?

TIA

Kevin B






From kevin1a at varlog.net  Mon Apr 26 22:32:38 2004
From: kevin1a at varlog.net (Kevin Brouelette)
Date: Mon, 26 Apr 2004 22:32:38 -0700
Subject: [OCLUG-devel] Learning awk
In-Reply-To: <1083042964.1490.20.camel@athlon.kblan.com>
References: <1083042964.1490.20.camel@athlon.kblan.com>
Message-ID: <1083043958.1490.23.camel@athlon.kblan.com>

On Mon, 2004-04-26 at 22:16, Kevin Brouelette wrote:
> Hello,
> 
> I'm spending my weeknights this week brushing up on bash shell.
> The book I'm working with is Addison-Wesley's 'Linux & Unix shell 
> Programing' by David Tansley.
> 
> Chapter 9 is about 'awk'. 
> After reading the chapter,
> I thought I would try to use awk to grab the users 
> which their UID from /etc/passwd is greater than 999.
> [Slackware UIDs start at 1000]
> 
> So I try this:
> awk -F: '{if($2 > 999) print$0"\t"$2}' /etc/passwd
> 
> hoping to print the name, a 'tab', and the UID.
> 
> It echo's the whole /etc/passwd file and no errors.
> Am I close or am I off base with trying to do this with awk?
> 
> TIA
> 
> Kevin B


The obvious strikes. $0 prints the whole line. 

awk -F: '{if($3 > 999) print$1"\t"$3}' /etc/passwd

Works.  I have a lot of reading to do. :)

Kevin 



From panfisher at earthlink.net  Tue Apr 27 10:33:31 2004
From: panfisher at earthlink.net (Tony Dycks)
Date: Tue, 27 Apr 2004 10:33:31 -0700
Subject: [OCLUG-devel] Learning awk/Another Decent Book
References: <1083042964.1490.20.camel@athlon.kblan.com>
	<1083043958.1490.23.camel@athlon.kblan.com>
Message-ID: <000d01c42c7d$c2b1c1a0$a536fea9@percival>


----- Original Message ----- 
From: "Kevin Brouelette" <kevin1a at varlog.net>
To: <oclug-devel at oclug.org>
Sent: Monday, April 26, 2004 10:32 PM
Subject: Re: [OCLUG-devel] Learning awk


> On Mon, 2004-04-26 at 22:16, Kevin Brouelette wrote:
> > Hello,
> >
> > I'm spending my weeknights this week brushing up on bash shell.
> > The book I'm working with is Addison-Wesley's 'Linux & Unix shell
> > Programing' by David Tansley.
> >
> > Chapter 9 is about 'awk'.
> > After reading the chapter,
> > I thought I would try to use awk to grab the users
> > which their UID from /etc/passwd is greater than 999.
> > [Slackware UIDs start at 1000]
> >
> > So I try this:
> > awk -F: '{if($2 > 999) print$0"\t"$2}' /etc/passwd
> >
> > hoping to print the name, a 'tab', and the UID.
> >
> > It echo's the whole /etc/passwd file and no errors.
> > Am I close or am I off base with trying to do this with awk?
> >
> > TIA
> >
> > Kevin B
>
>
> The obvious strikes. $0 prints the whole line.
>
> awk -F: '{if($3 > 999) print$1"\t"$3}' /etc/passwd
>
> Works.  I have a lot of reading to do. :)
>
> Kevin
>
>
>
Another good reference on Awk is the O'Reilly "Sed & Awk" publication by
Dale Dougherty & Arnold Robbines. Cover price; $29.95 for the 2nd Edition
(might be a newer version out)

Awk is a great shorthand shell language for working with consistent file
structures.


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