php

Question

我尝试在 m-d-Y 中增加天数 (60)，但计算错误。它被认为是 d-m-y 格式。

echo "Today is :: ".date("m-d-Y"); 
echo "\n";       
echo $DateEnd = date('m-d-Y', strtotime(date("m-d-Y") . ' +60 day'));

输出为::

Today is :: 11-07-2017
End date is :: 09-09-2017

代码：https://3v4l.org/jmT9a

如有任何帮助或建议，我们将不胜感激

Answer 1

问题原因：- strtotime() 注意部分：-

Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed. If, however, the year is given in a two digit format and the separator is a dash (-, the date string is parsed as y-m-d.

因此 m-d-Y 在向 php 日期添加天数时是无法识别的格式。

像下面一样：-

<?php

echo "Today is :: ".date("m-d-Y"); 
echo "\n";       
echo $DateEnd = date('m-d-Y', strtotime(date("y-m-d") . '+60 days'));

出坑：-https://eval.in/894300

同时检查可能的 Php date formats

Answer 2

阅读来自 PHP Manual 的注释：

Note: Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed. If, however, the year is given in a two digit format and the separator is a dash (-, the date string is parsed as y-m-d. To avoid potential ambiguity, it's best to use ISO 8601 (YYYY-MM-DD) dates or DateTime::createFromFormat() when possible.

所以你需要使用

echo date("m-d-Y");
echo "\n";
echo $contractDateEnd = date('m-d-Y', strtotime(date("m/d/Y") . ' +60 day'));

DEMO

Answer 3

试试这个，它会为你工作

<?php

echo "Current Date: ".date("m-d-Y");
echo "\n";
echo $contractDateEnd = date('m-d-Y', strtotime(date("d-m-Y") . ' +60 day'));
?>

Answer 4

您也可以使用日期间隔：

var_dump (date_create()->add(date_interval_create_from_date_string('60 days')));

Test at ideone

Answer 5

您的代码有几个问题：它获取当前时间（一个数字）并将当前日期格式化为字符串然后（在将“+60 天”连接到它之后）它再次将其解析为数字只是为了再次将其格式化为字符串。很多无用又耗电的操作。真正的麻烦是 custom date format 您用于日期的 PHP 出于充分的理由不支持它。

使用DateTime class 来处理日期和时间操作。它优于旧的 date()、strtotime() 和其他日期和时间函数。它可以处理多个时区，代码更短更清晰：

$now = new DateTime();
$end = new DateTime();
$end->add(new DateInterval('P60D'));

echo "Today is :: " . $now->format('m-d-Y') . "\n";
echo "End date is :: " . $end->format('m-d-Y') . "\n";

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