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);

它不起作用;问题是什么？

错误消息是：

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

您需要先使用a将日期字符串转换为DateTime对象自定义格式或可用的许多DateTime :: Format :: *库之一。您使用的是数据库中常用的格式，因此我选择了MySQL formatter（然后为最终结果定义自定义持续时间格式化程序，从中的示例复制而来 DateTime::Format::Duration）：

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。看到 DateTime。并且，您不能简单地打印返回值 subtract_datetime因为它返回一个引用。你必须使用方法，例如hours，用于提取有用的信息。

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

嗯，乍一看，我想如果你减去两个日期时间对象，$dt->subtract_datetime(...)将会起作用。

即：您的$date应该是日期时间

如何计算Perl中的DateTime差异？

3 个答案: