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. 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]
