CODEWITHSUNDEEP
phpleap-year
scorpdevil
scorpdevil

Reputation: 13

How to subtract a year from leap year

I'm looking for a solution for the following:

$date = new DateTime('02/29/2020');
$date->sub(new DataInterval('P1Y'));
$date->format('m/d/Y');

This Returns: 03/01/2019

Is there a way to return 02/28/2019? Thanks

Upvotes: 1

Views: 478

Answers (2)

delboy1978uk
delboy1978uk

Reputation: 12374

date('L') returns true if a leap year, so:

<?php

$date = new DateTime('02/29/2020');
$date->format('L') && $date->format('m-d') == '02-29' ? 'P366D' : 'P1Y';
$date->sub(new DateInterval($interval));
echo $date->format('m/d/Y');

Check it out here https://3v4l.org/4mXX4

Upvotes: 2

jspit
jspit

Reputation: 7703

The challenge is general: Add or subtract a month to a date without jumping to the next month if the date is close to the end of the month. For a year, this is to add or subtract 12 months.

The algorithm for the addMonthCut() function was taken here.

function addMonthCut(DateTime $date,$month = 1) {
  $dateAdd = clone $date;
  $dateLast = clone $date;
  $strAdd = ' '.(int)$month.' Month';
  $strLast = 'last Day of '.(int)$month.' Month';
  if ($dateAdd->modify($strAdd) < $dateLast->modify($strLast)) {
    $date->modify($strAdd);
  }
  else {
    $date->modify($strLast);
  }
  return $date;
}

Examples:

$date = new DateTime('02/29/2020');
$date = addMonthCut($date,-12);  //-1 Year
echo $date->format('m/d/Y');  //02/28/2019

$date = new DateTime('02/28/2019');
$date = addMonthCut($date,+12);  //+1 Year
echo $date->format('m/d/Y');  //02/28/2020

$date = new DateTime('01/31/2020');
$date = addMonthCut($date,+1);  //+1 Month
echo $date->format('m/d/Y');  //02/29/2020

Upvotes: 0

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