adding 1 day to a DATETIME format value

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?

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');

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);

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

1. Use strtotime to convert the string to a time stamp
2. 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.

You can use

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

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)));

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();

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.

There is a more concise and intuitive way to add days to php date. Don't get me wrong, those php expressions are great, but you always have to google how to treat them. I miss auto-completion facility for that.

Here is how I like to handle those cases:

(new Future(
    new DateTimeFromISO8601String('2014-11-21T06:04:31.321987+00:00'),
    new OneDay()
))
    ->value();

For me, it's way more intuitive and autocompletion works out of the box. No need to google for the solution each time.

As a nice bonus, you don't have to worry about formatting the resulting value, it's already is ISO8601 format.

This is meringue library, there are more examples here.