PHP, Get tomorrows date from date
PhpDateDatetimeStrtotimePhp Problem Overview
I have a PHP date in the form of 2013-01-22 and I want to get tomorrows date in the same format, so for example 2013-01-23.
How is this possible with PHP?
Php Solutions
Solution 1 - Php
Use DateTime
$datetime = new DateTime('tomorrow');
echo $datetime->format('Y-m-d H:i:s');
Or:
$datetime = new DateTime('2013-01-22');
$datetime->modify('+1 day');
echo $datetime->format('Y-m-d H:i:s');
Or:
$datetime = new DateTime('2013-01-22');
$datetime->add(new DateInterval("P1D"));
echo $datetime->format('Y-m-d H:i:s');
Or in PHP 5.4+:
echo (new DateTime('2013-01-22'))->add(new DateInterval("P1D"))
->format('Y-m-d H:i:s');
Solution 2 - Php
$tomorrow = date("Y-m-d", strtotime('tomorrow'));
or
$tomorrow = date("Y-m-d", strtotime("+1 day"));
Help Link: STRTOTIME()
Solution 3 - Php
Since you tagged this with [tag:strtotime], you can use it with the +1 day modifier like so:
$tomorrow_timestamp = strtotime('+1 day', strtotime('2013-01-22'));
That said, it's a much better solution to use DateTime.
Solution 4 - Php
<? php
//1 Day = 24*60*60 = 86400
echo date("d-m-Y", time()+86400);
?>
Solution 5 - Php
echo date ('Y-m-d',strtotime('+1 day', strtotime($your_date)));
Solution 6 - Php
Use DateTime:
To get tomorrow from now :
$d = new DateTime('+1day');
$tomorrow = $d->format('d/m/Y h.i.s');
echo $tomorrow;
>Results : 28/06/2017 08.13.20
To get tomorrow from a date :
$d = new DateTime('2017/06/10 08.16.35 +1day')
$tomorrow = $d->format('d/m/Y h.i.s');
echo $tomorrow;
>Results : 11/06/2017 08.16.35
Hope it helps!
Solution 7 - Php
/**
* get tomorrow's date in the format requested, default to Y-m-d for MySQL (e.g. 2013-01-04)
*
* @param string
*
* @return string
*/
public static function getTomorrowsDate($format = 'Y-m-d')
{
$date = new DateTime();
$date->add(DateInterval::createFromDateString('tomorrow'));
return $date->format($format);
}
Solution 8 - Php
By strange it can seem it works perfectly fine: date_create( '2016-02-01 + 1 day' );
echo date_create( $your_date . ' + 1 day' )->format( 'Y-m-d' );
Should do it
Solution 9 - Php
First, coming up with correct abstractions is always a key. key to readability, maintainability, and extendability.
Here, quite obvious candidate is an ISO8601DateTime. There are at least two implementations: first one is a parsed datetime from a string, and the second one is tomorrow. Hence, there are two classes that can be used, and their combination results in (almost) desired outcome:
new Tomorrow(new FromISO8601('2013-01-22'));
Both objects are an ISO8601 datetime, so their textual representation is not exactly what you need. So the final stroke is to make them take a date-form:
new Date(
new Tomorrow(
new FromISO8601('2013-01-22')
)
);
Since you need a textual representation, not just an object, you invoke a value() method.
For more about this approach, take a look at this post.
Solution 10 - Php
here's working function
function plus_one_day($date){
$date2 = formatDate4db($date);
$date1 = str_replace('-', '/', $date2);
$tomorrow = date('Y-m-d',strtotime($date1 . "+1 days"));
return $tomorrow; }
Solution 11 - Php
$date = '2013-01-22';
$time = strtotime($date) + 86400;
echo date('Y-m-d', $time);
Where 86400 is the # of seconds in a day.