Increment the day value from current day - php

Question

For example, in my case, i already has this value of array.....
array(1) {
[1]=>
array(1120) {
["2006-02-25"]=>
array(1) {
[0]=>
int(33)
}
["2006-02-20"]=>
array(1) {
[0]=>
int(38)
}
["2006-02-28"]=>
array(1) {
[0]=>
int(46)
}
that result I've got this code
$explodeEndDate = explode(" ",$adEndDate);
$explodeStartDate = explode(" ", $adStartDate);
$StartDate = $explodeStartDate[0];
$NewStartDate = strtotime("$explodeStartDate[0]");
$NewEndDate = strtotime("$explodeEndDate[0]");
$timeDiff = abs($NewEndDate - $NewStartDate);
// 86400 seconds in one day
$NumberDays = $timeDiff/86400;
//convert into int
$NumberDays = intval($NumberDays);
if(array_key_exists($NumberDays, $array[$itemType]) == false){
$array[$itemType][$StartDate] =[$NumberDays];
}
}
what I wanted to achieve is, in the "$StartDate" value which is for example ["2006-02-28"] , I wanted to plus it with the value of it. If we referring back to the figure above for example is
["2006-02-25"]=>
array(1) {
[0]=>
int(33)
so 2006-02-25 is the plus by 33 and the result is 2006-04-01. and after that, i wanted to make the date within that range

Answer 1

If you want to add date so here you go:
Months:
<?php
$date = date('Y-m-d', strtotime("+33 months", strtotime('2006-02-25')));
var_dump($date)
?>
Demo.
Days:
<?php
$date = date('Y-m-d', strtotime("+33 days", strtotime('2006-02-25')));
var_dump($date)
?>
Demo