Adding one day to a date

user_109859 0
Viewed 637036

My code to add one day to a date returns a date before day adding: 2009-09-30 20:24:00 date after adding one day SHOULD be rolled over to the next month: 1970-01-01 17:33:29

<?php

    //add day to date test for month roll over

    $stop_date = date('Y-m-d H:i:s', strtotime("2009-09-30 20:24:00"));

    echo 'date before day adding: '.$stop_date; 

    $stop_date = date('Y-m-d H:i:s', strtotime('+1 day', $stop_date));

    echo ' date after adding one day. SHOULD be rolled over to the next month: '.$stop_date;
?>

I have used pretty similar code before, what am I doing wrong here?

date datetime php
14 Answers

It Worked for me: For Current Date

$date = date('Y-m-d', strtotime("+1 day"));

for anydate:

date('Y-m-d', strtotime("+1 day", strtotime($date)));
user_59071570

The modify() method that can be used to add increments to an existing DateTime value.

Create a new DateTime object with the current date and time:

$due_dt = new DateTime();

Once you have the DateTime object, you can manipulate its value by adding or subtracting time periods:

$due_dt->modify('+1 day');

You can read more on the PHP Manual.

user_87349870 
<?php

function plusTimetoOldtime($Old_Time,$getFormat,$Plus_Time) {
    return date($getFormat,strtotime(date($getFormat,$Old_Time).$Plus_Time));
}

$Old_Time = strtotime("now");
$Plus_Time = '+1 day';
$getFormat = 'Y-m-d H:i:s';

echo plusTimetoOldtime($Old_Time,$getFormat,$Plus_Time);

?>
user_121923370

You can add one day to today's date using DateTime() and DateInterval() like this:

echo (new DateTime())->add(new DateInterval('P1D'))->format('Y-m-d');

OR a specific date:

$date = '2009-09-30 20:24:00';
echo (new DateTime($date))->add(new DateInterval('P1D'))->format('Y-m-d');
user_17674120

As we often receive ISO strings via an API from another time zone, we can make the conversion to local time on the fly:

// The machine local time is GMT+2, but we received a date at GMT+0 (UTC)
echo date(DATE_ISO8601, strtotime("-1 hour", strtotime("2022-08-17T23:25:51-00:00")));
// 2022-08-18T00:25:51+0200
user_24947540

Since you already have an answer to what's wrong with your code, I can bring another perspective on how you can play with datetimes generally, and solve your problem specifically.

Oftentimes you find yourself posing a problem in terms of solution. This is just one of the reasons you end up with an imperative code. It's great if it works though; there are just other, arguably more maintainable alternatives. One of them is a declarative code. The point is asking what you need, instead of how to get there.

In your particular case, this can look like the following. First, you need to find out what is it that you're looking for, that is, discover abstractions. In your case, it looks like you need a date. Not just any date, but the one having some standard representation. Say, ISO8601 date. There are at least two implementations: the first one is a date parsed from an ISO8601-formatted string (or a string in any other format actually), and the second is some future date which is a day later. Thus, the whole code could look like that:

(new Future(
    new DateTimeParsedFromISO8601('2009-09-30 20:24:00'),
    new OneDay()
))
    ->value();

For more examples with datetime juggling check out this one.

user_6180200
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