在Perl中如何找到给定日期的上一个星期一的日期？ [英] In Perl how to find the date of the previous Monday for a given date?

问题描述

我正在寻找一个Perl脚本，可以在任何指定的日期给我上个星期一。

I am looking for a Perl script which can give me the last Monday for any specified date.

对于日期2011-06-11，脚本应该返回2011-06-06

e.g. For date 2011-06-11, the script should return 2011-06-06

推荐答案

我假设如果给定日期是星期一，你想要同一个日期（而不是前一个星期一）。以下是 DateTime 的一种方法：

I'm assuming that if the given date is a Monday, you want the same date (and not the previous Monday). Here's one way to do it with DateTime:

use DateTime;

my $date = DateTime->new(year => 2011, month => 6, day => 11);
my $desired_dow = 1;            # Monday
$date->subtract(days => ($date->day_of_week - $desired_dow) % 7);
print "$date\n";

（实际上，对于星期一的特殊情况，％7 不是必需的，因为 $ date-> day_of_week - 1 将始终为0– 6，而mod 7是一个无效使用％7 ，它适用于任何所需的星期几，而不仅仅是星期一。）

(Actually, for the special case of Monday, the % 7 isn't necessary, because $date->day_of_week - 1 will always be 0–6, and that mod 7 is a no-op. But with the % 7, it works for any desired day-of-week, not just Monday.)

如果你想要上一个星期一，你可以改变减法：

If you did want the previous Monday, you can change the subtraction:

$date->subtract(days => ($date->day_of_week - $desired_dow) % 7 || 7);

如果您需要解析在命令行输入的日期，您可能需要查看 DateTime :: Format :: Natural 。

If you need to parse a date entered on the command line, you might want to look at DateTime::Format::Natural.

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