如何识别每一天所在行的日期？

Question

请看一看！

我有从1到30天的天数，所以我需要循环通过天数，并使用一个" for“循环确定每个对应日期的日期，开始日期是：$epoc = 2020-05-11；所以，我正在将其转换为epoc秒，我发现最后一个日期(第30天).I能够像这样解决它，但我需要一个"for”循环。有可能吗？

   #!/usr/bin/perl
   use Time::Local;
   use Time::Localtime;
   $day_a1 = 1;
   $day_a2 = 30;
   my ($year, $month, $day) = split('-', $epoc);
    $epoc = timelocal($s, $m, $h, $day, $month-1, $year-1900);
    $interval=$day_a2*60*60*24;
    $epoc1=$epoc+$interval;
    print scalar(localtime($epoc1)), "\n";

    @x1=(); 
    @date1=(); 
    for ($d = $day_a1; $d <= $day_a2; $d++){
            push (@x1, $d);
        }
    for ($d = $epoc; $d <= $epoc1; $d+=86400){
    print scalar(localtime($d));
    push (@date1, $d);
    }
    print @x1;
    print @date1;

Answer 1

如果你使用了可用的工具，你的生活就会变得容易得多。特别是，自2007年以来，Time::Piece和Time::Seconds一直是标准Perl发行版的一部分。

#!/usr/bin/perl

use strict;
use warnings;
use feature 'say';

use Time::Piece;
use Time::Seconds;

my $start = '2020-05-11';

# Add time of midday to avoid DST problems.
my $curr = Time::Piece->strptime("$start 12:00:00", '%Y-%m-%d %H:%M:%S');

for (1 .. 30) {
  say $curr->strftime('%Y-%m-%d');
  $curr += ONE_DAY;
}

Answer 2

也许您打算以下面的形式编写代码

#!/usr/bin/perl

use strict;
use warnings;
use feature 'say';

use Time::Local;
use Time::Localtime;

my $days = 30;
my $date = '2020-05-11';
my $day_sec = 60*60*24;
my ($year, $month, $day) = split('-', $date);
my $epoch = timelocal(0, 0, 0, $day, $month-1, $year-1900);

for(my $day=0; $day<$days; $day++) {
    say scalar localtime($epoch);
    $epoch += $day_sec;
}