adding 1 day to a DATETIME format value

Question

In certain situations I want to add 1 day to the value of my DATETIME formatted variable:

$start_date = date('Y-m-d H:i:s', strtotime("{$_GET['start_hours']}:{$_GET['start_minutes']} {$_GET['start_ampm']}"));

What is the best way to do this?

score 255 · Answer 1 · edited Feb 03 '17 at 16:16

255

There's more then one way to do this with DateTime which was introduced in PHP 5.2. Unlike using strtotime() this will account for daylight savings time and leap year.

$datetime = new DateTime('2013-01-29');
$datetime->modify('+1 day');
echo $datetime->format('Y-m-d H:i:s');

// Available in PHP 5.3

$datetime = new DateTime('2013-01-29');
$datetime->add(new DateInterval('P1D'));
echo $datetime->format('Y-m-d H:i:s');

// Available in PHP 5.4

echo (new DateTime('2013-01-29'))->add(new DateInterval('P1D'))->format('Y-m-d H:i:s');

// Available in PHP 5.5

$start = new DateTimeImmutable('2013-01-29');
$datetime = $start->modify('+1 day');
echo $datetime->format('Y-m-d H:i:s');

edited Feb 03 '17 at 16:16

Jasper

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answered Jan 31 '13 at 01:57

John Conde

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Ah, a *loop da loop*. – Funk Forty Niner Mar 12 '15 at 00:50
10

If adding more than one day, you can use either plural or singular and get the same result e.g.: `+2 day` or `+2 days` (obviously `+2 days` is more semantically correct) – timhc22 May 15 '15 at 12:57
1

What are the practical differences between these methods? The first example seems the simplest and most semantic to me. – BadHorsie Feb 13 '17 at 10:08

score 77 · Accepted Answer · answered Aug 17 '09 at 05:23

77

If you want to do this in PHP:

// replace time() with the time stamp you want to add one day to
$startDate = time();
date('Y-m-d H:i:s', strtotime('+1 day', $startDate));

If you want to add the date in MySQL:

-- replace CURRENT_DATE with the date you want to add one day to
SELECT DATE_ADD(CURRENT_DATE, INTERVAL 1 DAY);

answered Aug 17 '09 at 05:23

RaYell

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wont this give me the current time? i still need to keep my user inputted time. – ian Aug 17 '09 at 05:26
Yes it will but I marked you what values you should change to the custom time. You can just replace `time()` with `strtotime("{$_GET['start_hours']}:{$_GET['start_minutes']} {$_GET['start_ampm']}")`. – RaYell Aug 17 '09 at 05:37
this is not summer-/wintertime save! Please use DST save date/time manipulation – steven Jul 28 '17 at 10:35

score 37 · Answer 3 · edited Oct 19 '17 at 12:44

37

The DateTime constructor takes a parameter string time. $time can be different things, it has to respect the datetime format.

There are some valid values as examples :

'now' (the default value)
2017-10-19
2017-10-19 11:59:59
2017-10-19 +1day

So, in your case you can use the following.

$dt = new \DateTime('now +1 day'); //Tomorrow
$dt = new \DateTime('2016-01-01 +1 day'); //2016-01-02

edited Oct 19 '17 at 12:44

Anthony

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answered Sep 15 '16 at 10:20

Dmitry

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Answering without explanation? Read [How do I write a good answer?](http://stackoverflow.com/help/how-to-answer) – Shashanth Sep 15 '16 at 10:26

score 13 · Answer 4 · answered Aug 17 '09 at 05:27

13

Use strtotime to convert the string to a time stamp
Add a day to it (eg: by adding 86400 seconds (24 * 60 * 60))

eg:

$time = strtotime($myInput);
$newTime = $time + 86400;

If it's only adding 1 day, then using strtotime again is probably overkill.

answered Aug 17 '09 at 05:27

nickf

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So I need to convert it to a TIMESTAMP then back to a DATETIME format? – ian Aug 17 '09 at 05:32
4

I guess it should also be noted that not every day is 24 hours long. In most places in the world, daylight savings (summer time) means that one day a year is only 23 hours, and one other day is 25 hours. – nickf Jan 16 '12 at 10:21
1

Ups. You convert a datetime to a timestamp, so it's UTC based. And _UTC_ days are always 24h long. When converting back with `date()` the local timezone is taken into account which will take care of daylight saving. – rabudde Oct 25 '18 at 07:58

score 13 · Answer 5 · answered Aug 01 '19 at 10:43

13

You can use

$now = new DateTime();
$date = $now->modify('+1 day')->format('Y-m-d H:i:s');

answered Aug 01 '19 at 10:43

Omar

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score 7 · Answer 6 · answered Oct 29 '19 at 05:50

You can use as following.

$start_date = date('Y-m-d H:i:s');
$end_date = date("Y-m-d 23:59:59", strtotime('+3 days', strtotime($start_date)));

You can also set days as constant and use like below.

if (!defined('ADD_DAYS')) define('ADD_DAYS','+3 days');
$end_date = date("Y-m-d 23:59:59", strtotime(ADD_DAYS, strtotime($start_date)));

score 1 · Answer 7 · answered Aug 17 '09 at 05:39

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I suggest start using Zend_Date classes from Zend Framework. I know, its a bit offtopic, but I'll like this way :-)

$date = new Zend_Date();
$date->add('24:00:00', Zend_Date::TIMES);
print $date->get();

answered Aug 17 '09 at 05:39

Pawka

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score 1 · Answer 8 · answered Sep 19 '22 at 15:41

1

One liner !

echo (new \DateTime('2016-01-01 +1 day'))->format('Y-m-d H:i:s');

answered Sep 19 '22 at 15:41

Brain90

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score 0 · Answer 9 · answered Aug 25 '19 at 17:02

Using server request time to Add days. Working as expected.

25/08/19 => 27/09/19

$timestamp = $_SERVER['REQUEST_TIME'];
$dateNow = date('d/m/y', $timestamp);
$newDate = date('d/m/y', strtotime('+2 day', $timestamp));

Here '+2 days' to add any number of days.

adding 1 day to a DATETIME format value

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