为什么perl的本地时间似乎输出错误的月份？

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问题描述

我已经看到其他答案，但是由于某些原因他们并不适合我。

I've seen other answers, but they're not working for me for some reason.

我试图用Perl脚本中的日期以下代码。

I'm trying to get yesterday's date in a Perl script using the following code.

为了将来参考，今天的日期是2015年11月12日

For future reference, today's date is November 12, 2015

my (undef, undef, undef, $mday,$mon,$year) = localtime();
my $prev_day = timelocal(0, 0, 0, $mday-1, $mon, $year);
(undef, undef, undef, $mday,$mon,$year) = localtime($prev_day);
$year += 1900;

print "Yesterday's Day: $mday, Month: $mon, Year: $year\n";

除了我的输出看起来像这样

Except my output looks like this

Yesterday's Day: 11, Month: 10, Year: 2015.

我应该阅读昨天的日期为 Day：11，Month：11，Year：2015 。为什么要扣除这个月份？

I should be reading yesterday's date as Day: 11, Month: 11, Year: 2015. Why is the month being subtracted?

编辑：这不同于建议的答案，因为我想知道为什么当地时间似乎写错了一个月。 >

This is different than the suggested answer because I'm wondering why local time seems to be writing the wrong month.

推荐答案

的文档说这个

The documentation for localtime says this

所以要得到一个月的数字为1，您需要 $ mon + 1 ，或者您只需将一个添加到 $ mday 在你添加1900到$ code> $ year的同一时间

So to get a month number with 1 for January you need $mon+1, or you can just add one to $mday at the same time as you add 1900 to $year

像这样

use strict;
use warnings 'all';

use Time::Local;

my ($y, $m, $d) = (localtime)[5,4,3];

my $yesterday = timelocal(0, 0, 0, $d-1, $m, $y);
($y, $m, $d) = (localtime($yesterday))[5,4,3];
$y += 1900;
++$m;

print "Yesterday's Day: $d, Month: $m, Year: $y\n";

输出

output

Yesterday's Day: 11, Month: 11, Year: 2015

更好的方法是使用核心库 Time :: Piece 像这样

A better way is to use the core library Time::Piece like this

use strict;
use warnings 'all';

use Time::Piece;
use Time::Seconds 'ONE_DAY';

my $yesterday = localtime() - ONE_DAY;

printf "Yesterday's Day: %d, Month: %d, Year: %d\n",
    $yesterday->mday,
    $yesterday->mon,
    $yesterday->year;

输出

output

Yesterday's Day: 11, Month: 11, Year: 2015

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LocalTime