将 1 天添加到 DATETIME 格式值

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?

Answer 1

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');

Answer 2

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);

Answer 3

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

Answer 4

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.

Answer 5

You can use

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

Answer 6

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)));

Answer 7

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();

Answer 8

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.

Answer 9

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.

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

Answer 10

One liner !

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