perl如何将字符串转换为日期时间？

Question

我试图在 perl 中将字符串转换为日期，但出现错误。

use strict; 
use warnings;  
use DateTime;  
use Date::Manip;

my $date = ParseDate("20111121");
print "today is ".$date->day_of_week."\n"; 

错误

Can't call method "day_of_week" without a package or object reference 

看起来包导入有问题...

谢谢

Answer 1

DateTime 本身没有解析工具，但是有很多parsers生成 DateTime 对象。大多数时候，您可能需要DateTime::Format::Strptime .

use DateTime::Format::Strptime qw( );
my $format = DateTime::Format::Strptime->new(
   pattern   => '%Y%m%d',
   time_zone => 'local',
   on_error  => 'croak',
);
my $dt = $format->parse_datetime('20111121');

或者你可以自己做。

use DateTime qw( );
my ($y,$m,$d) = '20111121' =~ /^([0-9]{4})([0-9]{2})([0-9]{2})\z/
   or die;
my $dt = DateTime->new(
   year      => $y,
   month     => $m,
   day       => $d,
   time_zone => 'local',
);