Perl - 如何从给定的日期获取上一个星期三的日期，而不使用DateTime [英] Perl - How to get date of Previous wednesday from the given date without using DateTime

问题描述

全部，

我想从给定的日期找出上个星期三的日期。

I want to find out the date of previous wednesday from the given date.

例如。我有日期为20150804，我需要20150729。

Eg. I have date as "20150804" and i would need "20150729".

DateTime不可用，我也无法安装。

DateTime is not available and i cannot install it as well.

我看了几个例子，但是他们使用的是DateTime。

I looked at few examples but they were using DateTime.

你可以重定向我可以得到一些帮助吗？谢谢。

Can you please redirect me where i can get some help.? Thanks.

我打算在下面编写代码。

I am planning to code something like below.

代码：

#!/opt/perl-5.8.0/bin/perl

use warnings;
use strict;

my $dt="20150804";
my $prevWednesday=getPrevWednesday($dt);

sub getPrevWednesday()
{
    my $givenDt=shift;
    ...
}

推荐答案

另一个强力的方法，这次使用另一个核心模块 Time :: Local 。

Another brute force approach, this time using another core module Time::Local.

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

use Time::Local;

sub prev_wednesday {
    my $date = shift;
    my ($year, $month, $day) = $date =~ /(....)(..)(..)/;
    my $time = timelocal(0, 0, 12, $day, $month - 1, $year);
    do { $time -= 60 * 60 * 24 } until (localtime $time)[6] == 3; # <- Wednesday
    my ($y, $m, $d) = (localtime $time)[5, 4, 3];
    return sprintf "%4d%02d%02d\n", 1900 + $y, $m + 1, $d;
}

print $_, ' ', prev_wednesday($_), for qw( 20150804 20150805 20150806
                                           20150101 20000301 20010301 );

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