Question

如何在perl中转换1461241125.31307。我试过了：

use Date::Parse;
$unix_timestamp = '1461241125.31307';
my ($sec, $min, $hour, $mday, $mon, $year, $wday, $yday, $isdst) = localtime($unix_timestamp);
$mon += 1;
$year += 1900;
$unix_timestamp_normal = "$year-$mon-$mday $hour:$min:$sec";

结果：2016-4-21 5:18:45（没有填充小时）

如何填充它并使其成为GMT。我希望结果说2016-04-21 12:18:45

感谢大家的回答。

use DateTime;
$unix_timestamp = '1461241125.31307';
my $dt = DateTime->from_epoch(epoch => $unix_timestamp);
print $dt->strftime('%Y-%m-%d %H:%M:%S'),"\n";

Answer 1

最简单的方法：

print scalar localtime $unix_timestamp;

文档：http://perldoc.perl.org/functions/localtime.html

对于GMT，请使用gmtime：

print scalar gmtime $unix_timestamp;

文档：http://perldoc.perl.org/functions/gmtime.html（基本上说：localtime之类的所有内容，但输出GMT时间。）

对于自定义格式，请尝试DateTime：

use DateTime;

my $dt = DateTime->from_epoch(epoch => $unix_timestamp);
print $dt->strftime('%Y-%s');

有关所有选项，请参阅http://search.cpan.org/perldoc?DateTime。使用预定义的DateTime Formatters可以更轻松地创建大量格式：http://search.cpan.org/search?query=DateTime%3A%3AFormat&mode=all

Answer 2

use POSIX qw( strftime );

my $epoch_ts = '1461241125.31307';

say strftime('%Y-%m-%d %H:%M:%S', gmtime($epoch_ts));

Answer 3

使用gmtime代替localtime

perldoc -f gmtime：

gmtime EXPR
gmtime  Works just like "localtime" but the returned values are localized
        for the standard Greenwich time zone.

        Note: When called in list context, $isdst, the last value returned
        by gmtime, is always 0. There is no Daylight Saving Time in GMT.

        Portability issues: "gmtime" in perlport.

如何将unix纪元时间转换为perl中的可读格式

3 个答案: