The overwhelming majority submitted the following: awk '$3>=15{print $3}' Thanks to all those who responded. Regards, Mike > -----Original Message----- > From: Hickey, Michael > Sent: Tuesday, April 22, 2003 11:05 AM > To: Sun Managers Mailing List (E-mail) > Subject: Awk script help > > Can someone push me in the right direction of the proper syntax to take a > file, > dr.out containing multiple rows and 3 fields as below, and print only the > lines that > have a number >= 15 in field 3? > > 04/12/03 20:56 15.3 > 04/12/03 21:56 15.3 > 04/12/03 22:56 15.3 > 04/12/03 23:57 18.5 > > Thanks in advance. > > Mike _______________________________________________ sunmanagers mailing list sunmanagers@sunmanagers.org http://www.sunmanagers.org/mailman/listinfo/sunmanagersReceived on Wed Apr 23 07:58:53 2003
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