From: Zaigui Wang (zaigui@yahoo.com)
Date: Thu Jul 31 2003 - 16:52:33 EDT
The concensus seems to be just grep the patters from
the file and pipe it out to tail -1. I was hoping for
something fancier, but this will work...
Crist Clark also had this nice awk command:
awk '/<your date pattern>/{last=$0} END{print last}'
Thanks to:
rmck
Mark Scarborough
Frank Velazquez
David Foster
Crist Clark
Daniel Baldoni
Bochnik, William J
Janson.Santos@pinnaclewest.com
David.markowitz@sspsolutions.com
mike.salehi@kodak,com
Zaigui Wang wrote:
>
> Managers,
>
> I need to get the last line in a file that matches a
> date pattern (e.g. /03/07/03). How would I do that?
>
> sed seems to be a good choice. I was able to get a
> paricular line from a file this way with sed: sed -n
> 16p file, but I am open to all solutions that work.
>
> Any idea?
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