Re: date ranges as days

[punkish]

Thanks wind and ikegami. Once again, my apologies for this very simple problem that I should have figured out myself. I am adding my own solution to your contributions for the benefit of the archives

use Time::JulianDay;

my @start = (1981, 2, 1);
my @end   = (1982, 2, 5);
my $days = days('start' => \@start, 'end' => \@end, 'type' => 'yearly'
+);
my $days = days('start' => \@start, 'end' => \@end, 'type' => 'range')
+;

sub days {
    my (%a) = @_;
    
    my ($start, $end, $type) = ($a{'start'}, $a{'end'}, $a{'type'});
    
    if ($type eq 'range') {
        my $j0 = julian_day(1980, 1, 1);
        my $j1 = julian_day(@$start);
        my $j2 = julian_day(@$end);
        
        return (($j1 - $j0), ($j2 - $j0));
    }
    elsif ($type eq 'yearly') {
        my $start_year = $start->[0];
        my $end_year = $end->[0];
        
        my %days;
        for my $year ($start_year .. $end_year) {
            my $j0 = julian_day($year, 1, 1);
            my $j1; my $j2;
            if ($year == $start_year) {
                $j1 = julian_day($year, $start->[1], $start->[2]);
                $j2 = julian_day($year, 12, 31);
            }
            elsif ($year == $end_year) {
                $j1 = julian_day($year, 1, 1);
                $j2 = julian_day($year, $end->[1], $end->[2]);        
+    
            }
            else {
                $j1 = julian_day($year, 1, 1);
                $j2 = julian_day($year, 12, 31);
            }
            
            $days{"$year"} = [($j1 - $j0) .. ($j2 - $j0)];
        }
        
        return \%days;
    }
}