Question

在将它们转换为纪元后，我计算了两个ISO 8601日期的差异。如何在天数中得出它们的差异？我的代码是

my $ResolvedDate = "2014-06-04T10:48:07.124Z";
my $currentDate = "2014-06-04T06:03:36-04:00"

my $resolved_epoch = &convert_time_epoch($ResolvedDate);
my $current_epoch = &convert_time_epoch($currentDate);

if (($resolvedDate - $currentDate) > $noOfDays) {
    print "Difference in greater than x\n";
    $built = 0;
    return ($built);
} else {
    print "Difference in smaller than x \n";
    $built = 1;
    return ($built);
}

sub convert_time_epoch {
    my $time_c = str2time(@_);
    my @time_l = localtime($time_c);
    my $epoch = strftime("%s", @time_l);

    return($epoch);
}

除了$ built之外，我还要返回确切的天数，已解决的日期大于当前日期。

Answer 1

“天数”很尴尬，因为这是本地时间和DST存在（或者至少可能存在）。

通过简单地除以86400，您可以轻松获得24小时的数量，这可能足以达到您的目的。

如果您想要mday字段发生变化的真实次数，则可能与此简单除法所获得的值略有不同。

Answer 2

如果日期是在纪元秒中，请取差值并将其除以一天中的秒数（即86400）。像这样：

my $days_difference = int(($time1 - $time2) / 86400);

如果您使用DateTime，那么

my $duration = $dt1->delta_days($dt2); #$dt1 and $dt2 are DateTime objects.
print $duration->days;

Answer 3

use DateTime::Format::ISO8601 qw( );

my $ResolvedDate = "2014-06-04T10:48:07.124Z";
my $currentDate = "2014-06-04T06:03:36-04:00";

my $format = DateTime::Format::ISO8601->new();

my $dt_resolved = $format->parse_datetime($ResolvedDate);
my $dt_current  = $format->parse_datetime($currentDate);

my $dur = $dt_resolved->delta_days($dt_current);
my $days = $dur->in_units('days');

将纪元差异转换为天数

3 个答案: