如何在 PHP 中将日期增加一天？

Question

I am using this this syntax to increase one day above but when i put this format it still give me wrong date like this. '01/01/1970' But I want format and date like this '25/08/2016'.

$today = '24/08/2016';
$nextday = strftime("%d/%m/%Y", strtotime("$today +1 day"));

so please help me how can i do this.advance thanx.

Answer 1

You can use strtotime .

$your_date = strtotime("1 day", strtotime("2016-08-24"));
$new_date = date("Y-m-d", $your_date);

Hope it will help you.

Answer 2

It's important to note the different interpretation of - and / in the date. If you use a - php will determine it to be DD-MM , if you use a / php will determine it to be MM-DD . So you need to use - instead of /

<?php

$today = '24-08-2016';
echo $nextday = date("d-m-Y", strtotime("$today +1 day"));

?>

Answer 3

See below perfect working code

<?php
    echo $startDate = date('Y-m-d H:i:s'); echo "<br/>";
    echo $nextDate = date("Y-m-d H:i:s", strtotime("$startDate +1 day"));
?>

Answer 4

If you want to use DateTime:

$today      = '24/08/2016';    
$nextDay    = DateTime::createFromFormat('d/m/Y', $today)
                ->add(new DateInterval('P1D'))
                ->format('d/m/Y');

Answer 5

You should replace the / with -

Dates in the m/d/y or dmy formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/) , then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.) , then the European** dmy format is assumed**.

Reference: http://php.net/manual/en/function.strtotime.php

Try this:

$today = '24/08/2016';
$today = str_replace('/', '-', $today);
$today = date('Y-m-d', strtotime($today));
$nextday = date("d/m/Y", strtotime($today. "+1 day")); // Output: 25/08/2016

Answer 6

Please use below code

<?php
  $today = '24/08/2016';
  $today = explode('/',$today);
  $nextday = date('d/m/Y',mktime(0,0,0,$today[1],$today[0]+1,$today[2])));
?>