Increase current date by 5 days - php

Question

$date = date('Y-m-d',current_time('timestamp', 0));
How do I change $date to $date + 5 days?
PHP version is 5.2.
This code doesn't work:
$date_cur = date('Y-m-d',current_time('timestamp', 0));
echo $date_cur . ' <br>';
$date_cur_plus = date($date_cur, strtotime('+5 days'));
echo $date_cur_plus;
Gives me:
2011-11-29
2011-11-29

Answer 1

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

Answer 2

You could use mktime() using the timestamp.
Something like:
$date = date('Y-m-d', mktime(0, 0, 0, date('m'), date('d') + 5, date('Y')));
Using strtotime() is faster, but my method still works and is flexible in the event that you need to make lots of modifications. Plus, strtotime() can't handle ambiguous dates.
Edit
If you have to add 5 days to an already existing date string in the format YYYY-MM-DD, then you could split it into an array and use those parts with mktime().
$parts = explode('-', $date);
$datePlusFive = date(
'Y-m-d',
mktime(0, 0, 0, $parts[1], $parts[2] + 5, $parts[0])
// ^ Month ^ Day + 5 ^ Year
);

Answer 3

Object oriented Style:
<?php
$date = new DateTime('now');
$date->add(new DateInterval('P5D'));
echo $date->format('Y-m-d') . "\n";
?>
Procedural Style:
<?php
$date = date_create('2016-01-01');
date_add($date, date_interval_create_from_date_string('5 days'));
echo date_format($date, 'Y-m-d');
?>

Answer 4

Use strtotime:
$date = date('Y-m-d', strtotime('+5 days'));

Answer 5

$dateplus5 = date('Y-m-d', strtotime('+5 days'));

Answer 6

For specific date:
$date = '2011-11-01';
$date_plus = date('Y-m-d', strtotime($date.'+5 days'));
echo $date.'<br>'.$date_plus;
It will be give :
2011-11-01
2011-11-06

Answer 7

You can use
strtotime(“+5 days”)
to get the current date plus 5 days or
$targetDate = date($date, strtotime(’+5 days’));

Answer 8

strtotime() is very nice. It allows you to do the following:
$startDate = 'now'; // or choose a certain date you want/have
$startDate = '2013-02-14';
$intervals = array(
'',
'+ 5 days',
'+ 31 days',
'+ 3 months',
'+ 2 years + 2 days'
);
foreach($intervals as $interval) {
$combinedDate = $startDate . ' ' . $interval;
var_dump($combinedDate . ' => ' date('Y-m-d', strtotime($combinedDate)));
}
with a result:
now => 1360334498 = 2013-02-08
now + 5 days => 1360766498 = 2013-02-13
now + 31 days => 1363012898 = 2013-03-11
now + 3 months => 1368020498 = 2013-05-08
now + 2 years + 2 days => 1423579298 = 2015-02-10
or:
2013-02-14 => 1360792800 = 2013-02-14
2013-02-14 + 5 days => 1361224800 = 2013-02-19
2013-02-14 + 31 days => 1363471200 = 2013-03-17
2013-02-14 + 3 months => 1368478800 = 2013-05-14
2013-02-14 + 2 years + 2 days => 1424037600 = 2015-02-16

Answer 9

I used this:
$date = strtotime("+1 day", strtotime("2007-02-28"));
echo date("Y-m-d", $date);
It's working now.

Answer 10

Did not supposed to be like this?
$date_cur = date('Y-m-d', current_time('timestamp', 0));
echo $date_cur . ' <br>';
$date_cur_plus = date('Y-m-d', strtotime('+5 days', current_time('timestamp', 0) ) );
echo $date_cur_plus;

Answer 11

if the date is already exists, you can used this code :
$tartDate = date("m/d/Y", strtotime("+1 Day", strtotime($Date)));