On 12/21/2010 1:58 PM, Roland RoLaNd wrote: > > the only thing missing is to find a way to just take the earliest time of each day. > > in other words the above output should be: > > > 0 DaysLate > That means my perl script was wrong... This looks more like what you want, except for your last change to 9:10. my %id_count; my %id_date; #date already seen; my %iddate_time; #1st time each day while (<>) { my ($x,$id,$date,$time,$junk) = split /,/; next if ($x == 'X'); #skip header $iddate_time{$id . $date} = $time unless ($iddate_time{$id . $date}); #store earliest next if ($time le "09:00:00"); # not late $t = $iddate_time{$id . $date}; next if ($iddate_time{$id . $date} le "09:00:00"); # 1st wasn't late next if ($id_date{$id} eq $date); # already counted today print "Late: $id - $date - $time\n"; $id_count{$id}++; $id_date{$id} = $date; } print "----\n"; while (( my $id,$count) = each(%id_count)) { print "$id late $count days\n"; } -- Les Mikesell lesmikesell@xxxxxxxxx _______________________________________________ CentOS mailing list CentOS@xxxxxxxxxx http://lists.centos.org/mailman/listinfo/centos