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Top Forums Shell Programming and Scripting shell script to remove all lines which exceeds a particular date & time Post 302502537 by panyam on Tuesday 8th of March 2011 08:40:58 AM
Old 03-08-2011
For your sample file, I got the output as below.

Code:
 
$ awk -F"[ :]" '$6~"03/08/11"&&$7>=6' sample_file
6 ixtarchive-b IXT Memphis_Prod_SQL_Diff Memphis-Prod-SQL-Inc-Application-Backup 03/08/11 06:32:58
150 pbnaaedb004-b Scratch Airpark_Prod_SQL_Diff_1 Airpark-Prod-SQL-Inc-Application-Backup 03/08/11 11:06:57

 

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PACEMAKER(8)						  System Administration Utilities					      PACEMAKER(8)

NAME
Pacemaker - Part of the Pacemaker cluster resource manager SYNOPSIS
iso8601 command [output modifier] DESCRIPTION
iso8601 - Display and parse ISO8601 dates and times OPTIONS
-?, --help This text -$, --version Version information -V, --verbose Increase debug output Commands: -n, --now Display the current date/time -d, --date=value Parse an ISO8601 date/time. Eg. '2005-01-20 00:30:00 +01:00' or '2005-040' -p, --period=value Parse an ISO8601 date/time with interval/period (wth start time). Eg. '2005-040/2005-043' -D, --duration=value Parse an ISO8601 date/time with duration (wth start time). Eg. '2005-040/P1M' Output Modifiers: -L, --local Show result as a 'local' date/time -O, --ordinal Show result as an 'ordinal' date/time -W, --week Show result as an 'calendar week' date/time For more information on the ISO8601 standard, see: http://en.wikipedia.org/wiki/ISO_8601 AUTHOR
Written by Andrew Beekhof REPORTING BUGS
Report bugs to pacemaker@oss.clusterlabs.org Pacemaker 1.1.7 April 2012 PACEMAKER(8)
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