perl - 我如何在Perl中获得本周的日期？

Question

我有以下循环来计算当前星期的日期并打印出来。它可以工作，但是我在Perl中花了很多时间查看日期/时间，并想就是否有更好的方法征询您的意见。这是我编写的代码:

#!/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 );
}

如前所述，这可行，但是最快还是最有效？我猜想速度和效率可能不一定会结合在一起，但您的反馈意见值得我们赞赏。

Answer 1

弗里多的答案的略有改进的版本...

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。对于星期日，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方法而不是像Friedo那样重新实现它；)