Help with dates

mhearse has asked for the wisdom of the Perl Monks concerning the following question:

I've made a script to fetch files via Net::FTP. The files are in directories which correspond with days of the week (i.e. Sun Mon Tue etc...). When fetching the files I need to rename them like this
FILE-11-12-03
Using the current date, which would correspond with the day of the week directory. For example: the file in /Wed would be named FILE-11-12-03. The file in /Tue would be named FILE-11-11-03, and so on. I'm sure there must be an easy way (maybe some module), but I'm oblivious. Thanks.

use Net::FTP;

die "Please specify site.\n" unless ($ARGV[0]);
$site = $ARGV[0];
chomp ($today = `date '+%a'`);
chomp ($d = `date '+%d'`);
chomp ($m = `date '+%m'`);
chomp ($y = `date '+%y'`);
$d -= 1;
@days = grep { $_ ne $today } qw/Sun Mon Tue Wed Thu Fri Sat/;
%month = (1, 31, 2, 29, 3, 31, 4, 30, 5, 31, 6, 30, 7, 31,
          8, 31, 9, 30, 10, 31, 11, 30, 12, 31);

$ftp = Net::FTP->new("$site", Debug => 0)
   or die "Cannot connect to $site: $@";

$ftp->login("reap",'only.hd')
   or die "Cannot login ", $ftp->message;

foreach $day (@days) {
   $ftp->get("/data/logfiles/old/$day/24HourSummary.txt" ,"$site.$m-$d
+-$y")
      or die "get failed ", $ftp->message;
   print "Finished $site.$m-$d-$y for $day.\n";
   $d -= 1;
   if ($d == 0) {
      $m -= 1;
      $m = 12 if $m = 0;
      $d = $month{$m};
   }
}

$ftp->quit;
[download]

Comment on Help with dates Download Code

Replies are listed 'Best First'.
Re: Help with dates by shockme (Chaplain) on Nov 12, 2003 at 23:48 UTC
Just use localtime: `my ($mday, $mon, $year, $wday) = (localtime)[3..6]; $mon += 1; $year += 1900;` [download] Of course, this leaves you with a 4-digit year, but that's easy enough to fix: `$year =~ s/^\d\d//;` [download] See `perldoc -f localtime` for more information. Update: Modified to include $wday and `perldoc` reference. If things get any worse, I'll have to ask you to stop helping me.	[reply] [d/l] [select]
Re: Re: Help with dates by thelenm (Vicar) on Nov 12, 2003 at 23:59 UTC
Of course, this leaves you with a 4-digit year, but that's easy enough to fix: You mean that's easy enough to unfix. :-) I don't know why people still use 2-digit years, but I guess if that's what the OP wants... -- Mike -- `XML::Simpler does not require XML::Parser or a SAX parser. It does require File::Slurp. -- grantm, perldoc XML::Simpler`	[reply]
Re: Re: Re: Help with dates by shockme (Chaplain) on Nov 13, 2003 at 00:03 UTC
I'm with you, and I originally wasn't even going to address the issue. But the OP's specs call for a 2-digit year, and far be it from me to violate specs. ;) If things get any worse, I'll have to ask you to stop helping me.	[reply]
Re: Help with dates by Paulster2 (Priest) on Nov 13, 2003 at 01:19 UTC
Check out the Date::Manip module on CPAN, or try perldoc as it has quite an extensive writeup on it. It will enable you to manipulate any date in any way that you want. chomp ($today = `date '+%a'`); chomp ($d = `date '+%d'`); chomp ($m = `date '+%m'`); chomp ($y = `date '+%y'`); [download] and `@days = grep { $_ ne $today } qw/Sun Mon Tue Wed Thu Fri Sat/; %month = (1, 31, 2, 29, 3, 31, 4, 30, 5, 31, 6, 30, 7, 31, 8, 31, 9, 30, 10, 31, 11, 30, 12, 31);` [download] With Date::Manip this part of the code will become obsolete. You will no longer need to fill the `%month` hash, as you can just check against it. You can also see which day of the week it is. Date::Manip also uses UNIX "+%" (ie: +%j for julian date). Strong mojo for the needy date coder. Paulster2	[reply] [d/l] [select]
Re: Help with dates by Ninthwave (Chaplain) on Nov 12, 2003 at 23:53 UTC
From the Perl Cookbook `use Date::Calc qw(Day_of_Week Week_Number Day_of_Year); # you have $year, $month, and $day # $day is day of month by definition $wday = Day_of_Week($year, $month, $day); $wnum = Week_Number($year, $month, $day); $dnum = Day_of_Year($year, $month, $day);` [download] Update I read that backwords expected numeric to name. So how would we reverse it hmmmm.... I would go with shockme because depending on how deep you have to go back your function mishandles leap years. "No matter where you go, there you are." BB	[reply] [d/l]