Simplest way to increment a date in PHP?

Question

Say I have a string coming in, "2007-02-28", what's the simplest code I could write to turn that into "2007-03-01"? Right now I'm just using strtotime(), then adding 24*60*60, then using date(), but just wondering if there is a cleaner, simpler, or more clever way of doing it.

score 86 · Accepted Answer · answered Mar 18 '09 at 23:41

86

A clean way is to use strtotime()

$date = strtotime("+1 day", strtotime("2007-02-28"));
echo date("Y-m-d", $date);

Will give you the 2007-03-01

answered Mar 18 '09 at 23:41

Ólafur Waage

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score 36 · Answer 2 · answered Mar 18 '09 at 23:35

36

It's cleaner and simpler to add 86400. :)

The high-tech way is to do:

$date = new DateTime($input_date);
$date->modify('+1 day');
echo $date->format('Y-m-d');

but that's really only remotely worthwhile if you're doing, say, a sequence of transformations on the date, rather than just finding tomorrow.

answered Mar 18 '09 at 23:35

chaos

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Stupid CentOS only has PHP 5.1, and DateTime is introduced in 5.2. I guess I finally have to upgrade to an unofficial centos php package then. – davr Mar 18 '09 at 23:45
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Er, no. Just adding 86400 is false simplicity. – staticsan Mar 19 '09 at 04:35
Could you explain what you mean by that? – chaos Mar 19 '09 at 04:52
It is quite easy to understand what 24*60*60 means at a glance. 86400 less so. – Tom Haigh Mar 19 '09 at 11:08
Ah. I guess I've been doing this too long, then. 86400 is as recognizable to me as 65536. – chaos Mar 19 '09 at 13:28
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@chaos: If your smallest unit of working time is a day, then you can usually get away with adding or substracting units of 86400. If it is smaller, you need to be aware of daylight savings and time zones. In which case, use the built-in functions. They will get it right more often than you will. – staticsan Mar 23 '09 at 22:16
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Not all days are 86400 seconds long (leap seconds). – Jimmy T. Sep 04 '16 at 13:26

score 17 · Answer 3 · edited Mar 19 '09 at 00:06

17

You can do the addition right inside strtotime, e.g.

 $today="2007-02-28";
 $nextday=strftime("%Y-%m-%d", strtotime("$today +1 day"));

edited Mar 19 '09 at 00:06

chaos

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answered Mar 18 '09 at 23:42

Paul Dixon

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Oh, neat. Didn't know that. It's almost like good old Date::Manip. – chaos Mar 19 '09 at 00:06

score 6 · Answer 4 · answered Jan 18 '15 at 15:20

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The simplest way...

echo date('Y-m-d',strtotime("+1 day"));    //from today

OR from specified date...

echo date('Y-m-d',strtotime("+1 day", strtotime('2007-02-28')));

answered Jan 18 '15 at 15:20

Biswadeep Sarkar

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score 6 · Answer 5 · edited Jan 03 '12 at 13:53

6

Another way is to use function mktime(). It is very useful function...

$date = "2007-02-28";
list($y,$m,$d)=explode('-',$date);
$date2 = Date("Y-m-d", mktime(0,0,0,$m,$d+1,$y));

but I think strtotime() is better in that situation...

edited Jan 03 '12 at 13:53

mac

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answered Mar 19 '09 at 00:48

Bajlo

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nice, I didn't realise that mktime() would increment the month like that when you go over – Tom Haigh Mar 19 '09 at 11:10

score 0 · Answer 6 · answered Jul 18 '17 at 21:11

Hello you can try this below especially if you are french

$date = date('l j F Y');
#increment the date
$date2 = date('l j F Y', strtotime("+7 day"));

to translate in french you can use the setlocale() function or the function below :

function fr_date($date){

    $date = explode(' ', $date);

    $date = str_replace('Monday','Lundi',$date);
    $date = str_replace('Tuesday','Mardi',$date);
    $date = str_replace('Wednesday','Mercredi',$date);
    $date = str_replace('Thursday','Jeudi',$date);
    $date = str_replace('Friday','Vendredi',$date);
    $date = str_replace('Saturday','Samedi',$date);
    $date = str_replace('Sunday','Dimanche',$date);

    $date = str_replace('January','Janvier',$date);
    $date = str_replace('February','Février',$date);
    $date = str_replace('March','Mars',$date);
    $date = str_replace('April','Avril',$date);
    $date = str_replace('May','Mai',$date);
    $date = str_replace('June','Juin',$date);
    $date = str_replace('July','Juillet',$date);
    $date = str_replace('August','Août',$date);
    $date = str_replace('September','Septembre',$date);
    $date = str_replace('October','Octobre',$date);
    $date = str_replace('November','Novembre',$date);
    $date = str_replace('December','Décembre',$date);

    $date = implode(' ',$date);
    return $date;

}

score 0 · Answer 7 · answered Oct 20 '17 at 01:03

0

$your_date = strtotime("1month", strtotime(date("Y-m-d")));
 $new_date = date("Y-m-d", $your_date++);

answered Oct 20 '17 at 01:03

Eatsleeprace Livelife

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/("1month")= will increment by month. //("1day")= will increment by day. //(date("Y-m-d"))=start to increment date on you current date. – Eatsleeprace Livelife Oct 20 '17 at 01:04
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You should edit your answer to include the explanation of the code, instead of having the explanation in a comment. – astidham2003 Oct 20 '17 at 01:12

score 0 · Answer 8 · answered Mar 18 '21 at 10:56

0

use strtotime() With date Formate

echo date('Y-m-d', strtotime('2007-02-28' . ' +1 day'));

answered Mar 18 '21 at 10:56

Paresh Shiyal

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ranojan · Answer 9 · 2017-04-04T09:09:00.407

-1

 $early_start_date = date2sql($_POST['early_leave_date']);


        $date = new DateTime($early_start_date);
        $date->modify('+1 day');


        $date_a = new DateTime($early_start_date . ' ' . $_POST['start_hr'] . ':' . $_POST['start_mm']);
        $date_b = new DateTime($date->format('Y-m-d') . ' ' . $_POST['end_hr'] . ':' . $_POST['end_mm']);

        $interval = date_diff($date_a, $date_b);

edited Apr 04 '17 at 09:09

answered Sep 04 '16 at 12:24

ranojan

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Simplest way to increment a date in PHP?

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