blocks|key|2885060|text|use+DateTime::Format::Strptime+qw(+);

my+$format+=+DateTime::Format::Strptime->new(
+++pattern++=>+'%25m/%25d/%25Y',
+++on_error+=>+'croak',
);

my+$dt+=+$format->parse_datetime('10/14/2013');
$dt->subtract(+months+=>+4+);
say+$format->format_datetime($dt);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2885061|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>use DateTime::Format::Strptime qw( );

my $format = DateTime::Format::Strptime-&gt;new(
 pattern =&gt; '%m/%d/%Y',
 on_error =&gt; 'croak',
);

my $dt = $format-&gt;parse_datetime('10/14/2013');
$dt-&gt;subtract( months =&gt; 4 );
say $format-&gt;format_datetime($dt);
</code></pre>

blocks|key|568668|text|$+date+'%2B%25m/%25d/%25Y'+--date='02/14/2013+4+month+ago'
10/14/2012|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|568669|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>$ date '+%m/%d/%Y' --date='02/14/2013 4 month ago'
10/14/2012
</code></pre>

blocks|key|878993|text|CPAN+module+Date::Manip非常适合这类事情。Date::Manip非常强大，因为它可以理解工作日和假日...并且可以解析任意的日期字符串，如“下周二”或“八月的第三个星期四”。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|878994|你应该试一试，我打赌你会发现它很有用。|878995|下面是一个简短的示例，它可以解决您当前的问题：|878996|#!/usr/bin/perl

use+Date::Manip;

my+$d+=+new+Date::Manip::Date("10/18/2013");
print+$d->printf("%25c"),+"\n";
my+$delta+=+new+Date::Manip::Delta("4+months+ago");
my+$od+=+$d->calc($delta);
print+$od->printf("%25c"),+"\n";|code-block|syntax|javascript|878997|...and这将生成以下输出：|878998|Fri+Oct+18+00:00:00+2013
Tue+Jun+18+00:00:00+2013|878999|entityMap|0|LINK|mutability|MUTABLE|url|http://search.cpan.org/~sbeck/Date-Manip-6.41/lib/Date/Manip/Calc.pod^0|0|N|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|C|$]]|$1|O|3|P|5|J|7|16|8|@]|9|@]|C|$K|L]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

The <a href="http://search.cpan.org/~sbeck/Date-Manip-6.41/lib/Date/Manip/Calc.pod" rel="nofollow">CPAN module Date::Manip</a> is perfect for this sort of thing. Date::Manip is quite powerful in that it understands business days and holidays... and can parse arbitrary date strings like "next Tuesday" or "3rd Thursday in August".

You should give it a shot; I bet you'll find it quite useful.

Here's a short example the solves your current problem:

<pre><code>#!/usr/bin/perl

use Date::Manip;

my $d = new Date::Manip::Date("10/18/2013");
print $d-&gt;printf("%c"), "\n";
my $delta = new Date::Manip::Delta("4 months ago");
my $od = $d-&gt;calc($delta);
print $od-&gt;printf("%c"), "\n";
</code></pre>

...and this generates the following output:

<pre><code>Fri Oct 18 00:00:00 2013
Tue Jun 18 00:00:00 2013
</code></pre>

blocks|key|2885116|text|当心那些有更多天数的月份！|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2885117|下面是一个例子，减去一个月得到的答案可能是你意想不到的|2885118|endOfMarch=$(date+-d+"2017-03-31")
today=$endOfMarch
echo+$(date+-d+"$today+-1+month"+"%2B%25Y-%25m-%25d")

2017-03-02|code-block|syntax|javascript|2885119|因为二月只有28天，减去一个月实际上就产生了日期2017-02-31，这个日期显然不存在，因此已经溢出到给出2017-03-02的三月的额外天数|CODE|2885120|根据您试图解决的问题，您可以通过创建一个设置为当前月份的第一天的日期变量，然后减去月份，可靠地计算出几个月前的n。|2885121|endOfMarch=$(date+-d+"2017-03-31")
today=$endOfMarch
firstDayOfMonth=$(date+-d+"$today"+"%2B%25Y-%25m-01")
echo+$(date+-d+"$firstDayOfMonth+-1+month"+"%2B%25Y-%25m-%25d")

2017-02-01|2885122|因此，现在您有了正确的Month+(即减去1个月后的2月)，然后您必须根据您的用例确定将新日期设置为哪个Day。|2885123|因此，增加/减少月份并不像人们想象的那样简单|2885124|entityMap^0|0|D|0|0|0|O|A|1I|A|0|1J|1|0|0|B|5|1G|3|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|14|8|@$9|15|A|16|B|O]|$9|17|A|18|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|19|8|@$9|1A|A|1B|B|O]]|D|@]|E|$]]|$1|R|3|S|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|T|3|U|5|6|7|1D|8|@$9|1E|A|1F|B|O]|$9|1G|A|1H|B|O]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1I|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|Y|$]]

Beware of months that have more days in them than others!

Here's an example where subtracting one month gives you an answer that you are probably not expecting

<pre><code>endOfMarch=$(date -d "2017-03-31")
today=$endOfMarch
echo $(date -d "$today -1 month" "+%Y-%m-%d")

2017-03-02
</code></pre>

Because February only has 28 days, subtracting one month has effectively created the date <code>2017-02-31</code> which obviously doesn't exist and so has spilled over the extra days into March given you <code>2017-03-02</code>

Depending on what you are trying to solve you can reliably calculate <code>n</code> months ago, by creating a date variable set to the first day of the current month, then subtracting the months.

<pre><code>endOfMarch=$(date -d "2017-03-31")
today=$endOfMarch
firstDayOfMonth=$(date -d "$today" "+%Y-%m-01")
echo $(date -d "$firstDayOfMonth -1 month" "+%Y-%m-%d")

2017-02-01
</code></pre>

So now you have the correct <code>Month</code> (i.e. February when subtracting 1 month) and you then have to figure out what <code>Day</code> to set the new date to based on your use case.

So adding/subtracting Months is not as straight forward as one might think

I will be reading random dates from a files, I want to substract 4 months from that date. I am programming in Perl but I can use a bash command.

If i have the 10/14/2013 i will get 06/14/2013.

Thanks a lot!

Karem

Month ago from a date given a date in bash or perl

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我将从文件中读取随机日期，我想从该日期减去4个月。我是用Perl编程的，但我可以使用bash命令。如果我有2013年10月14日，我将获得2013年6月14日。非常感谢!卡雷姆

问从bash或perl中给定的日期开始的一个月前
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从bash或perl中给定的日期开始的一个月前EN