如何在Perl中计算DateTime差值？

Question

use DateTime ;

my $date = "2010-08-02 09:10:08";

my $dt = DateTime->now( time_zone => 'local' )->set_time_zone('floating');
print $dt->subtract_datetime($date);

它不工作；有什么问题吗？

错误消息为：

Can't call method "time_zone" without a package or object reference at
/opt/perl/perl5.12/lib/site_perl/5.12.0/x86_64-linux/DateTime.pm line 1338

Answer 1

您需要首先使用一种自定义格式或许多可用的DateTime:: format ::*库中的一个将日期字符串转换为DateTime对象。您使用的是数据库中常用的格式，所以我选择了MySQL格式化程序(然后为最终结果定义了一个自定义的持续时间格式化程序，复制自DateTime::Format::Duration中的示例)：

use DateTime;
use DateTime::Format::MySQL;
use DateTime::Format::Duration;

my $date = "2010-08-02 09:10:08";

my $dt1 = DateTime->now(time_zone => 'floating', formatter => 'DateTime::Format::MySQL');
my $dt2 = DateTime::Format::MySQL->parse_datetime($date);

my $duration = $dt1 - $dt2;
my $format = DateTime::Format::Duration->new(
    pattern => '%Y years, %m months, %e days, %H hours, %M minutes, %S seconds'
);
print $format->format_duration($duration);

# prints:
# 0 years, 00 months, 0 days, 00 hours, 421 minutes, 03 seconds

Answer 2

$date必须是DateTime对象，而不是简单的字符串。参见DateTime。而且，您不能简单地打印subtract_datetime的返回值，因为它返回一个引用。您必须使用hours等方法来提取有用的信息。

use strict;
use warnings;
use DateTime;

my $dt2 = DateTime->new(
                       year   => 2010,
                       month  => 8,
                       day    => 2,
                       hour   => 9,
                       minute => 10,
                       second => 8,
                       time_zone => 'local',
                     );

my $dt1 = DateTime->now( time_zone => 'local' )->set_time_zone('floating');

my $dur = $dt1->subtract_datetime($dt2);
print 'hours = ', $dur->hours(), "\n";

__END__

hours = 2

Answer 3

乍一看，我猜如果减去两个datetime对象，$dt->subtract_datetime(...)就可以工作了。

ie :你的$date应该是datetime