PHP DateTime createFormat from invalid date

Question

Hi I want to create DateTime object from date 1989-06-31. As you can see this date has bad count of days(June can't have 31 days). But calling this

$semi_valid_date = "1989-06-31";
$semi_valid_datetime = DateTime::createFromFormat('Y-m-d',$semi_valid_date);
$valid_date = $semi_valid_datetime->format('Y-m-d');

will store to $valid_date date 1989-07-01.

Is it possible to throw error, when month has more days, than it can have?

6th month or aka June doesn't have 31 days. That's way first next day is selected after 06-30, which is 07-01. — Glavić
– Glavić, Commented Jan 30, 2014 at 20:57

John Conde · Accepted Answer · 2014-01-30 19:23:23Z

1

Use checkdate() to validate the date before attempting to use it:

var_dump(checkdate(6, 31, 1989));
// bool(false)

var_dump(checkdate(6, 30, 1989));
// bool(true)

See it in action

answered Jan 30, 2014 at 19:23

John Conde

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