PHP日期将当前日期添加5年

时间:2013-08-16 09:00:13

标签: php

我有这个PHP代码:

$end=date('Y-m-d');

我用它来获取当前日期,我需要将来5年的日期,例如:

$end=date('(Y + 5)-m-d');

我该怎么做?

12 个答案:

答案 0 :(得分:123)

尝试:

$end = date('Y-m-d', strtotime('+5 years'));

答案 1 :(得分:18)

修改日期 based on this post
strtotime()非常强大,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

<强>日期时间

    $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

你可以在strtotime()投掷的东西是相当令人惊讶和非常人性化的。请看下周二寻找星期二的这个例子
程序

    $t = strtotime("Tuesday next week");
    echo date('r', $t) . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100

<强>日期时间

    $dt = new DateTime("Tuesday next week");
    echo $dt->format('r') . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100

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构造函数采用的完整时间格式列表列在PHP Supported Date and Time Formats page上。

另一个适合您案例的例子可能是: based on this post

    <?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));
    ?>

答案 2 :(得分:10)

使用此代码将特定日期的年,月或日或小时或分钟或秒添加

 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

您也可以减去替换+到 -

答案 3 :(得分:3)

       $date = strtotime($row['timestamp']);
       $newdate = date('d-m-Y',strtotime("+1 year",$date));

答案 4 :(得分:2)

使用碳非常容易。 $date = "2016-02-16"; // Or Your date $newDate = Carbon::createFromFormat('Y-m-d', $date)->addYear(1);

答案 5 :(得分:1)

要在今天的日期添加一年,请使用以下内容:

$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));

答案 6 :(得分:1)

您可以为此使用 DateInterval;

$currentDate = new \DateTime(); //creates today timestamp
$currentDate->add(new \DateInterval('P5Y')); //this means 5 Years
and you can now format it;
$currentDate->format('Y-m-d');

答案 7 :(得分:0)

使用Carbon

$dt = Carbon::now();
echo $dt->addYears(5); 

答案 8 :(得分:0)

试试这个,

$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;

答案 9 :(得分:0)

试试这个:

$yearnow= date("Y");
$yearnext=$yearnow+1;
echo date("Y")."-".$yearnext;

答案 10 :(得分:0)

尝试使用此代码并添加下几天,几个月和几年

// 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>';
}

答案 11 :(得分:0)

试试下面的代码,希望对你有帮助

<?php

$current_date=strtotime(date('Y-m-d'));

echo $end = date('Y-m-d', strtotime('+5 years',$current_date));

?>