PHP zero out the time for a date and add 3 days

Question

I'm looking for an elegant/efficient way to take out the time portion of a datetime in format 'Ymd H:i:s', and then add 3 days.

Currently the solution is:

date('Y-m-d 00:00:00', strtotime("+3 days", strtotime('2017-01-23 05:32:12')));

where 2017-01-23 05:32:12 is the date, and this correctly outputs 2017-01-26 00:00:00 .

It just feels like there has to be a better way to do this.

Thanks

Answer 1

DateTime() offers several of ways to do this. None of them are any less verbose than your current method:

// Plain old DateTime()
$date = (new DateTime('2017-01-23 05:32:12'))->modify('+3 days')->format('Y-m-d 00:00:00');

// DateTme using DateInterval to add three days
$date = (new DateTime('2017-01-23 05:32:12'))->add(new DateInterval('P3D'))->format('Y-m-d 00:00:00');

// DateTime setting the date to midnight instead of using 00:00:00
$date = (new DateTime('2017-01-23 05:32:12'))->modify('+3 days')->modify('midnight')->format('Y-m-d H:i:s');

If the date is today you can shorten this a bit:

$date = (new DateTime('+3 days'))->format('Y-m-d 00:00:00');

Answer 2

you can use DateTime class

$date = new DateTime('2017-01-23 05:32:12');
$date->modify('+3 days');
$outputDateString = $date->format('Y-m-d 00:00:00');

one line:

$date = ( new DateTime('2017-01-23 05:32:12'))->modify('+3 days')->format('Y-m-d 00:00:00');