如何更改Perl中日期时间值的时区?

基于UTC,epochtime只有一个标准定义,而不同时区没有不同的epochtime。

如果你想找到 offset between gmtime and localtime 使用

use Time::Local;
@t = localtime(time);
$gmt_offset_in_seconds = timegm(@t) - timelocal(@t);

use DateTime;
my $dt   = DateTime->now;
$dt->set_time_zone( 'Europe/Madrid' );

你只需要设定时区。尝试:

env TZ=UTC+1 perl -e 'use Time::Local; print timelocal("00","00","00","01","01","2000"),"\n";'

Time::Local::timelocal 是的反义词 localtime .结果将出现在主机的本地时间中:

$ perl -MTime::Local -le \
    'print scalar localtime timelocal "00","00","00","01","01","2000"'
Tue Feb  1 00:00:00 2000

你想要 gmtime 与之相对应的 本地时间 ?

$ perl -MTime::Local' -le \
    'print scalar gmtime timelocal "00","00","00","01","01","2000"'
Mon Jan 31 23:00:00 2000

你想换个方向吗? 本地时间 与之相对应的 通用时间 是吗?

$ perl -MTime::Local -le \
    'print scalar localtime timegm "00","00","00","01","01","2000"'
Tue Feb  1 01:00:00 2000

虽然时间::本地是一个合理的解决方案,但最好使用更现代的面向对象的datetime模块。下面是一个例子:

use strict;
use DateTime;
my $dt = DateTime->now;
print $dt->epoch, "\n";

对于时区,可以使用datetime::timezone模块。

use strict;
use DateTime;
use DateTime::TimeZone;

my $dt = DateTime->now;
my $tz = DateTime::TimeZone->new(name => "local");

$dt->add(seconds => $tz->offset_for_datetime($dt));

print $dt->epoch, "\n";

CPAN链接:

DateTime

另一个基于 DateTime::Format::Strptime

use strict;
use warnings;
use v5.10;
use DateTime::Format::Strptime;

my $s = "2016-12-22T06:16:29.798Z";
my $p = DateTime::Format::Strptime->new(
  pattern => "%Y-%m-%dT%T.%NZ",
  time_zone => "UTC"
);

my $dt = $p->parse_datetime($s);    
$dt->set_time_zone("Europe/Berlin");
say join ' ', $dt->ymd, $dt->hms; # shows 2016-12-22 07:16:29