我怎样才能在Perl中获得本周的日期？

我有以下循环来计算当前周的日期并打印出来.它有效,但我在Perl的日期/时间可能性游泳,想要了解是否有更好的方法.这是我写的代码：

#!/usr/bin/env perl
use warnings;
use strict;

use DateTime;

# Calculate numeric value of today and the 
# target day (Monday = 1,Sunday = 7); the
# target,in this case,is Monday,since that's
# when I want the week to start
my $today_dt = DateTime->Now;
my $today = $today_dt->day_of_week;
my $target = 1;

# Create DateTime copies to act as the "bookends"
# for the date range
my ($start,$end) = ($today_dt->clone(),$today_dt->clone());

if ($today == $target)
{
  # If today is the target,"start" is already set;
  # we simply need to set the end date
  $end->add( days => 6 );
}
else
{
  # Otherwise,we calculate the Monday preceeding today
  # and the Sunday following today
  my $delta = ($target - $today + 7) % 7;
  $start->add( days => $delta - 7 );
  $end->add( days => $delta - 1 );
}

# I clone the DateTime object again because,for some reason,# I'm wary of using $start directly...
my $cur_date = $start->clone();

while ($cur_date <= $end)
{
  my $date_ymd = $cur_date->ymd;
  print "$date_ymd\n";
  $cur_date->add( days => 1 );
}

如上所述,这是有效的,但它是最快还是最有效的？我猜测速度和效率可能不一定在一起,但您的反馈非常感谢.

解决方法

friedo的答案略有改进……

my $start_of_week =
    DateTime->today()
            ->truncate( to => 'week' );

for ( 0..6 ) {
    print $start_of_week->clone()->add( days => $_ );
}

但是,这假设星期一是一周的第一天.星期天,从…开始

my $start_of_week =
    DateTime->today()
            ->truncate( to => 'week' )
            ->subtract( days => 1 );

无论哪种方式,最好使用truncate方法而不是重新实现它,就像friedo所做的那样;)