Date: Tue, 6 Mar 2007 08:53:49 -0800 From: Gary Kline <kline@tao.thought.org> To: Derek Ragona <derek@computinginnovations.com> Cc: Gary Kline <kline@tao.thought.org>, FreeBSD Mailing List <freebsd-questions@freebsd.org> Subject: Re: awk question Message-ID: <20070306165349.GA67829@thought.org> In-Reply-To: <6.0.0.22.2.20070306072709.02577448@mail.computinginnovations.com> References: <20070306003506.GA12553@thought.org> <6.0.0.22.2.20070306072709.02577448@mail.computinginnovations.com>
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On Tue, Mar 06, 2007 at 07:27:56AM -0600, Derek Ragona wrote:
> You can loop through them using a shell script:
> for i in `ls -lt | awk '{if ($8 == 2006) print $9}'`;do rm $i;done
This is the safest way to rm or rm -i each file ($i); the
ls -ls | [awkstuff] spits out the entire list in one chunk.
But since packages from 2006 were OLD, I just /bin/rm'd them
wholesale. Thanks,
gary
>
> -Derek
>
>
> At 06:35 PM 3/5/2007, Gary Kline wrote:
>
> > Guys,
> >
> > Having found $9 , how do I /bin/rm it (using system()--yes??)
> > in an awk one-liner?
> >
> > I'm trying to remove from packages from long ago and find and
> > print them with
> >
> > ls -lt | awk '{if ($8 == 2006) print $9}';
> >
> > but what I want to remove the file pointed at by $9. I've tried
> > FILE=ARGV[9]; and using FILE within my system() call, but no-joy.
> > What's the magic here?
> >
> > thanks in advance,
> >
> > gary
> >
> >
> >
> >--
> > Gary Kline kline@thought.org www.thought.org Public Service Unix
> >
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--
Gary Kline kline@thought.org www.thought.org Public Service Unix
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