perl如何将字符串转换为Datetime？

Question

我试图在perl中将字符串转换为日期,但是会出错.

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,但有许多解析器可以生成DateTime对象.大多数情况下,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',
);

Answer 2

DateTime不解析日期.我会选择Time :: Piece核心模块给你strptime():

#!/usr/bin/env perl
use strict;
use warnings;
use Time::Piece;
my $t = Time::Piece->strptime("20111121", "%Y%m%d");
print $t->strftime("%w\n");