PHP date add 5 year to current date

Question

I have this PHP code:

$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:

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

How can I do this?

Answer 1

Try with:

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

Answer 2

Modifying dates based on this post
strtotime() is really powerful and allows you to modify/transform dates easily with it’s relative expressions too:

Procedural

    $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

    $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

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

DateTime

    $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.

Another example, suitable for your case could be: 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));
    ?>

Answer 3

Use this code to add years or months or days or hours or minutes or seconds to a given date

 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 -

Answer 4

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

Answer 5

Its very very easy with Carbon.

$date = "2016-02-16"; // Or Your date
$newDate    = Carbon::createFromFormat('Y-m-d', $date)->addYear(1);

Answer 6

Using Carbon:

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

Answer 7

To add one year to todays date use the following:

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

Answer 8

You may use DateInterval for this purpose;

$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');

Answer 9

Try this code:

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

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

?>

Answer 10

try this ,

$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;

Answer 11

try this:

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

Answer 12

Try this code and add next Days, Months and Years

// 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>';
}