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.

English

Answer 1

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

Lagos

Answer 2

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.

Timetable

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. – English

Er, no. Just adding 86400 is false simplicity. – Monochromatism

Could you explain what you mean by that? – Timetable

It is quite easy to understand what 24*60*60 means at a glance. 86400 less so. – Elliot

Ah. I guess I've been doing this too long, then. 86400 is as recognizable to me as 65536. – Timetable

@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. – Monochromatism

Not all days are 86400 seconds long (leap seconds). – Wembley

Answer 3

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

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

Copyhold

Oh, neat. Didn't know that. It's almost like good old Date::Manip. – Timetable

Answer 4

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...

Thaumaturgy

nice, I didn't realise that mktime() would increment the month like that when you go over – Elliot

Answer 5

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')));

Bookseller

Answer 6

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;

}

Vinnievinnitsa

Answer 7

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

Kenyakenyatta

/("1month")= will increment by month. //("1day")= will increment by day. //(date("Y-m-d"))=start to increment date on you current date. – Kenyakenyatta

You should edit your answer to include the explanation of the code, instead of having the explanation in a comment. – Pikeperch

Answer 8

use strtotime() With date Formate

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

Caravaggio

Answer 9

 $early_start_date = date2sql($_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);

Kantianism