在 PHP 中为日期添加天数
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Add days to a date in PHP
提问by user1038814
Is there any php function available where I can add days to a date to make up another date? For example, I have a date in the following format: 27-December-2011
是否有任何可用的 php 函数，我可以在其中添加天数以组成另一个日期？例如，我有以下格式的日期：27-December-2011
If I add 7 to the above, it should give: 03-January-2012.
如果我在上面加上 7，它应该给出：03-January-2012。
Many thanks
非常感谢
回答by Fabrizio
Try this
尝试这个
$add_days = 7;
$date = date('Y-m-d',strtotime($date) + (24*3600*$add_days));
回答by Simone
Look at this simple snippet
看看这个简单的片段
$date = date("Y-m-d");// current date
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 day");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 week");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +2 week");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +1 month");
$date = strtotime(date("Y-m-d", strtotime($date)) . " +30 days");
回答by Aurelio De Rosa
回答by Smamatti
date('Y-m-d', strtotime('+6 days', strtotime($original_date)));
回答by Lucian Minea
Actually it's easier than all that.
事实上，这比这一切都容易。
$some_var = date("Y-m-d",strtotime("+7 day"))
You can use a variable instead of the string, of course. It will be great if the people answering the questions, won't complicate things. Less code, means less time to waste on the server ;).
当然，您可以使用变量代替字符串。如果回答问题的人不会使事情复杂化，那就太好了。更少的代码，意味着更少的时间浪费在服务器上；)。
回答by Matthew
$date = new DateTime('27-December-2011');
$date->add(new DateInterval('P7D'));
echo $date->format('d-F-Y') . "\n";
Change the format string to be whatever you want. (See the documentation for date()).
将格式字符串更改为您想要的任何内容。（请参阅 的文档date()）。
回答by Wasim Khan
$registered = $udata->user_registered;
$registered = date( "d m Y", strtotime( $registered ));
$challanexpiry = explode(' ', $registered);
$day = $challanexpiry[0];
$month = $challanexpiry[1];
$year = $challanexpiry[2];
$day = $day+10;
$bankchallanexpiry = $day . " " . $month . " " . $year;
