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? __________________________________ Do you Yahoo!? Yahoo! SiteBuilder - Free, easy-to-use web site design software http://sitebuilder.yahoo.com _______________________________________________ sunmanagers mailing list sunmanagers@sunmanagers.org http://www.sunmanagers.org/mailman/listinfo/sunmanagersReceived on Thu Jul 31 16:56:04 2003
This archive was generated by hypermail 2.1.8 : Thu Mar 03 2016 - 06:43:17 EST