PHP 日期将 5 年添加到当前日期
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PHP date add 5 year to current date
提问by Letter
I have this PHP code:
我有这个 PHP 代码：
$end=date('Y-m-d');
I use it to get the current date, and I need the date 5 years in the future, something like:
我用它来获取当前日期，我需要未来 5 年的日期，例如：
$end=date('(Y + 5)-m-d');
How can I do this?
我怎样才能做到这一点？
回答by hsz
Try with:
尝试：
$end = date('Y-m-d', strtotime('+5 years'));
回答by Gadoma
Modifying datesbased on this post
strtotime() is really powerful and allows you to modify/transform dates easily with it's relative expressions too:
Procedural
基于这篇文章修改日期
strtotime() 非常强大，并且允许您使用它的相对表达式轻松修改/转换日期：
程序
$dateString = '2011-05-01 09:22:34';
$t = strtotime($dateString);
$t2 = strtotime('-3 days', $t);
echo date('r', $t2) . PHP_EOL; // returns: Thu, 28 Apr 2011 09:22:34 +0100
DateTime
约会时间
$dateString = '2011-05-01 09:22:34';
$dt = new DateTime($dateString);
$dt->modify('-3 days');
echo $dt->format('r') . PHP_EOL; // returns: Thu, 28 Apr 2011 09:22:34 +0100
The stuff you can throw at strtotime() is quite surprising and very human readable. Have a look at this example looking for Tuesday next week.
Procedural
您可以在 strtotime() 中抛出的内容非常令人惊讶并且非常易于阅读。看看这个例子，寻找下周的星期二。
程序
$t = strtotime("Tuesday next week");
echo date('r', $t) . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100
DateTime
约会时间
$dt = new DateTime("Tuesday next week");
echo $dt->format('r') . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100
Note that these examples above are being returned relative to the time now. The full list of time formats that strtotime() and the DateTime constructor takes are listed on the PHP Supported Date and Time Formats page.
请注意，上面的这些示例是相对于现在的时间返回的。strtotime() 和 DateTime 构造函数采用的时间格式的完整列表列在PHP 支持的日期和时间格式页面上。
Another example, suitable for your case could be:based on this post
另一个适合您情况的示例可能是：基于这篇文章
<?php
//How to get the day 3 days from now:
$today = date("j");
$thisMonth = date("n");
$thisYear = date("Y");
echo date("F j Y", mktime(0,0,0, $thisMonth, $today+3, $thisYear));
//1 week from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth, $today+7, $thisYear));
//4 months from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth+4, $today, $thisYear));
//3 years, 2 months and 35 days from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth+2, $today+35, $thisYear+3));
?>
回答by RN Kushwaha
Use this code to add years or months or days or hours or minutes or seconds to a given date
使用此代码将年、月、日、小时、分钟或秒添加到给定日期
echo date("Y-m-d H:i:s", strtotime("+1 years", strtotime('2014-05-22 10:35:10'))); //2015-05-22 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 months", strtotime('2014-05-22 10:35:10')));//2014-06-22 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 days", strtotime('2014-05-22 10:35:10')));//2014-05-23 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 hours", strtotime('2014-05-22 10:35:10')));//2014-05-22 11:35:10
echo date("Y-m-d H:i:s", strtotime("+1 minutes", strtotime('2014-05-22 10:35:10')));//2014-05-22 10:36:10
echo date("Y-m-d H:i:s", strtotime("+1 seconds", strtotime('2014-05-22 10:35:10')));//2014-05-22 10:35:11
You can also subtract replacing + to -
您还可以减去替换 + 到 -
回答by Nirav
$date = strtotime($row['timestamp']);
$newdate = date('d-m-Y',strtotime("+1 year",$date));
回答by Bijaya Prasad Kuikel
Its very very easy with Carbon.
$date = "2016-02-16"; // Or Your date
$newDate = Carbon::createFromFormat('Y-m-d', $date)->addYear(1);
使用碳非常容易。
$date = "2016-02-16"; // Or Your date
$newDate = Carbon::createFromFormat('Y-m-d', $date)->addYear(1);
回答by Deepu
To add one year to todays date use the following:
要将一年添加到今天的日期，请使用以下命令：
$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));
回答by Abhi
try this ,
尝试这个 ，
$presentyear = '2013-08-16 12:00:00';
$nextyear = date("M d,Y",mktime(0, 0, 0, date("m",strtotime($presentyear )), date("d",strtotime($presentyear )), date("Y",strtotime($presentyear ))+5));
echo $nextyear;
回答by GoldTechnia Pvt. Ltd.
try this:
尝试这个：
$yearnow= date("Y");
$yearnext=$yearnow+1;
echo date("Y")."-".$yearnext;
回答by Rahul Gandhi
Try this code and add next Days, Months and Years
试试这个代码并添加接下来的天、月和年
// current month: Aug 2018
$n = 2;
for ($i = 0; $i <= $n; $i++){
$d = strtotime("$i days");
$x = strtotime("$i month");
$y = strtotime("$i year");
echo "Dates : ".$dates = date('d M Y', "+$d days");
echo "<br>";
echo "Months : ".$months = date('M Y', "+$x months");
echo '<br>';
echo "Years : ".$years = date('Y', "+$y years");
echo '<br>';
}
