Re: Help

Another thing to find out would be how your putting the date into the data file. That's something we would need to know to figure out the "prior date" portion.

I can't come up with a quick non-CPAN way to do that to cover Dec 31 -> Jan 1 changes, and also lastday(month)->firstday(nextmonth) changes.

Any folks have any ideas on how to parse the date, assuming they're stored as MM/DD/YYYY (if they're separated fields, they can be join()'d, and if DD/MM/YYYY or some other format, it should be easily changable to MM/DD/YYYY)? Dates are probably one of the most irritating things when it comes to programming. =]

-marius

Comment on Re: Help

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(jeffa) Re: Re: Help by jeffa (Bishop) on Dec 06, 2000 at 22:06 UTC
I think this will do the trick: `use strict; use Time::Local; my $time = timelocal(localtime); $time -= (60 * 60 * 24); # subtract one day my ($s, $min, $h, $d, $mon, $y, $w, $yd, $isdst) = localtime($time); printf("Yesterday's date: %02d:%02d:%02d-%04d/%02d/%02d\n", $h, $min, $s, $y+1900, $mon+1, $d);` [download] The trick is to work in Epoch seconds, and let localtime figure out what the date is. Jeff L-LL-L--L-LL-L--L-LL-L-- -R--R-RR-R--R-RR-R--R-RR F--F--F--F--F--F--F--F-- (the triplet paradiddle)	[reply] [d/l]

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(jeffa) Re: Re: Help
by jeffa (Bishop) on Dec 06, 2000 at 22:06 UTC

use strict;
use Time::Local;

my $time = timelocal(localtime);
$time -= (60 * 60 * 24);          # subtract one day

my ($s, $min, $h, $d, $mon, $y, $w, $yd, $isdst) = localtime($time);
printf("Yesterday's date: %02d:%02d:%02d-%04d/%02d/%02d\n",
    $h, $min, $s, $y+1900, $mon+1, $d);
[download]

localtime

Jeff

L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
F--F--F--F--F--F--F--F--
(the triplet paradiddle)

[reply]
[d/l]