Paul Kraus wrote: > No your not missing anything I am :) I have never seen this before. > > Looking at the perl book it says that it will subtract a$ from the > highest multiple of b$ that is not greater then a$. > > So writing a quick loop that does 6 % $i where $i is 1..24 I get > these results. Which makes sense. But then how could I use this > operator to test for 456 , 10 11 12 ect.
You have the operation backwards. Try: $ perl -le 'print $_, " -> ", ($_ - 1) % 6 > 2 for 1..24' 2 -> 3 -> 4 -> 1 5 -> 1 6 -> 1 7 -> 8 -> 9 -> 10 -> 1 11 -> 1 12 -> 1 13 -> 14 -> 15 -> 16 -> 1 17 -> 1 18 -> 1 19 -> 20 -> 21 -> 22 -> 1 23 -> 1 24 -> 1 > > Even this test produces a 0 on six. Maybe I am missing something but I > don't see how this would work then again I have never seen this > before. Thanks for your patients. > > Paul > > 1 0 > 2 0 > 3 0 > 4 2 > 5 1 > 6 0 > 7 .. Infinity = 6. > > > > -----Original Message----- > From: Wagner, David --- Senior Programmer Analyst --- WGO > [mailto:[EMAIL PROTECTED] > Sent: Tuesday, July 08, 2003 10:52 AM > To: [EMAIL PROTECTED]; [EMAIL PROTECTED] > Subject: RE: Looping with multiples? > > > Paul Kraus wrote: > > I have a variable amount of text files that need to get parsed. The > > files come in groups of 6. I only need to parse files 4 5 and 6. So > > if > > > I had 3 groups waiting to be processed then I would have to read > > through and parse 4,5,6,10,11,12,16,17,18. > > > > The file names look like this... > > > > 6.1.200_(1)_(9443).txt > > Date_element of group_time.txt > > > > So I would processing all of the files for 6.1. > > Also I couldn't figure out a way to sort them easily as you will see > > me in my code so any suggestions there would be helpful also. > > > > How can I setup a loop that will run the correct amount of times and > > only parse these files from the group. Here is my code. > > > > > > #!/usr/bin/perl > > > > use warnings; > > use strict; > > use DBI; > > > > opendir ( DH, "." ) or die ( "Could not open $!\n" ); my @allfiles > > = readdir(DH); my %sales; > > foreach ( @allfiles ) { > > next unless /^\d+\.\d+\.\d+_/; > > push (@{$sales{$1}}, $_) if /^(\d+\.\d+\.\d+)_/; > > } > > > > foreach (sort keys %sales){ > > my $salesjournals = scalar(@{$sales{$_}}); > > &processdate($sales{$_}, $salesjournals); > > die; > > } > > > > sub processdate { > > my ($filenames, $salesjournals) = ( shift, shift ); > > my (%pel, %cfb, %leimkuehler, @files); > > foreach ( @ { $filenames } ){ > > push (@files, $1 . "-$_") if /_\((\d+)\)_/; > > } > > @files = sort {$a <=> $b} @files; > > foreach (@files){ > > my $tmp = $1 if s/^(\d+)-//; > > if ( $tmp == ){ > > print "$tmp \t $_\n"; > > } > > } > > } > > If you are only after the files which are defined by _(n)_ and > you only want the 4,5 and 6 th elements, then you should be able to > do a modulus 6 (ie, n % 6) and where 4,5,6 then process otherwise > bypass. If I am missing the obvious, sorry, but if going in a > numeric sequence, the modulo should do it for you. > > Wags ;) > > > ********************************************************** > This message contains information that is confidential > and proprietary to FedEx Freight or its affiliates. > It is intended only for the recipient named and for > the express purpose(s) described therein. > Any other use is prohibited. > **************************************************************** > > > -- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > > > -- > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
