Question

我想获得自纪元以来给定的微秒数，以本地时间与gmtime之间的小时数表示。到目前为止，我有这样的事情

# Microseconds since Epoch 
my $msec = 555329743301750;

# Convert to seconds 
my $sec = $msec/1000000;

my $val = (POSIX::mktime(localtime $sec) - POSIX::mktime(gmtime $sec)) / 60 / 60;

print "$val\n";

1）我得到的输出是-6。（CST本地时间）但是我期望-5。我期望的行为类似于运行bash命令的结果

`date -d 20190514 "+%z"`;

2）长话短说，如何计算与perl中的date -d 20190514 "+%z"类似的偏移量？

Answer 1

使用核心Time::Piece：

use strict;
use warnings;
use Time::Piece;

# Microseconds since Epoch 
my $usec = 555329743301750;

# Convert to seconds 
my $sec = $usec/1000000;

my $offset = localtime($sec)->tzoffset / 3600;

如果按照我在your other question上显示的方式构造它们，

Time :: Moment和DateTime也可以返回此值。

my $offset = $time_moment->offset / 60;

my $offset = $datetime->offset / 3600;

您也可以从日期创建这些对象，但是它当然会为您提供当天的午夜（当地时间）的偏移量。

use Time::Piece;
my $time = localtime->strptime('20190514', '%Y%m%d');

use Time::Moment;
use Role::Tiny ();
my $class = Role::Tiny->create_class_with_roles('Time::Moment', 'Time::Moment::Role::TimeZone');
my $time = $class->new(year => 2019, month => 5, day => 14)->with_system_offset_same_local;

use DateTime;
my $time = DateTime->new(year => 2019, month => 5, day => 14, time_zone => 'local');

Answer 2

这将为您提供与Linux“％z”格式相同的[+-] HHMM格式：

use POSIX qw[mktime];

my $tz = (localtime time)[8] * 60 - mktime(gmtime 0) / 60;
$tz = sprintf "%+03d%02d\n", $tz / 60, abs($tz) % 60;

如果您只想以小时为单位，则可以执行以下操作：

my $tz_offset_hours = (localtime time)[8] - mktime(gmtime 0)) / 3600;

奖金：：以下子例程将返回带有时区偏移和微秒的完整时间戳，如“ YYYY-MM-DD HH：MM：SS.nnnnnn [+-] HHMM”：< / p>

use POSIX qw[mktime strftime];
use Time::HiRes qw[gettimeofday];

sub timestamp () {
  my @now = gettimeofday;
  my $tz  = (localtime $now[0])[8] * 60 - mktime(gmtime 0) / 60;
  my $ts  = strftime("%Y-%m-%d %H:%M:%S", localtime $now[0]);
  return sprintf "%s.%06d %+03d%02d", $ts, $now[1], $tz / 60, abs($tz) % 60;
}

Perl：当从自考虑到夏令时的时代开始的秒数内输入时间时，获取gmtime与本地时间之间的偏移

2 个答案: