How to get year + 1 if the date is already passed

Question

$mpsAgeT = $today->diff($mpsAgeT);

Where $today is:

object(DateTime)[37]
  public 'date' => string '2019-06-17 13:40:56.888563' (length=26)
  public 'timezone_type' => int 3
  public 'timezone' => string 'Europe/Sofia' (length=12)

and $mpsAgeT is:

object(DateTime)[38]
  public 'date' => string '2016-06-15 13:40:56.000000' (length=26)
  public 'timezone_type' => int 3
  public 'timezone' => string 'Europe/Sofia' (length=12)

Returns me 3 years, but 15.06 is an older date, so I want to return 4 years, 3 are passed and 4 year is started.

If I just added 1 year, it will not work if date is not passed or equal.

For example:

17.06.2019 and 17.06.2016 - should returns me 3
17.06.2019 and 19.06.2016 - should returns me 3
17.06.2019 and 15.06.2016 - should returns me 4

Possible duplicate of [How to calculate the difference between two dates using PHP?](https://stackoverflow.com/questions/676824/how-to-calculate-the-difference-between-two-dates-using-php) — Dave, Jun 17 '19 at 11:20
So, add one unless the other properties on `$mpsAgeT` are 0? If `$today` is '2019-06-14' it'll return 2 and you expect 3, right ? — msg, Jun 17 '19 at 11:20

msg · Accepted Answer · 2019-06-17T13:14:38.133

As I suggested in the comments, you always have to add one, unless the dates are exactly equal to get your expected result:

$diff = $today->diff($mpsAgeT);
$years = $diff->y;

 if (addOne($diff)) {
   $years++;
 }

function addOne($interval) {
    // Iterate precision properties of Interval object
    // In this case, month and days, but can add hours, minutes (check edits)
    // If it's anything other than exact years, should add one
    $props = ['m', 'd'];
    foreach ($props as $prop) {
        // This returns as soon as a property is not 0
        if ($interval->{$prop} !== 0)
          return true;
    }
    return false;
}

Demo

Alternate solution without using Intervals:

// Find year difference
$years = $today->format('Y') - $mpsAgeT->format('Y');
// Add to original date (sets it to the same date of the current year)
// Only add if the date has passed
if ($today > $mpsAgeT->modify('+' . $years . ' years')) {
   $years++;
}

Demo

Works pefect, thank you, please add some explantation about `addOne` function for better answer : )) — MorganFreeFarm, Jun 17 '19 at 12:57

score 1 · Answer 2 · answered Jun 17 '19 at 11:36

You can simply check if all other variables are not 0

$today = new DateTime('2019-06-17 13:40:56.888563');
$mpsAgeT = new DateTime('2016-06-17 13:40:56.888563');
    $interval = $today->diff($mpsAgeT);
    $diff = $interval->format('%y');
    if (!($interval->d === 0 && $interval->m === 0 && $interval->h === 0 && $interval->i === 0 && $interval->s === 0)) {
        $diff++;
    }
    echo $diff; // returns 3

$today = new DateTime('2019-06-17 13:40:56.888563');
$mpsAgeT = new DateTime('2016-06-19 13:40:56.888563');
$interval = $today->diff($mpsAgeT);
$diff = $interval->format('%y');
if (!($interval->d === 0 && $interval->m === 0 && $interval->h === 0 && $interval->i === 0 && $interval->s === 0)) {
    $diff++;
}
echo $diff;// returns 3

$today = new DateTime('2019-06-17 13:40:56.888563');
$mpsAgeT = new DateTime('2016-06-15 13:40:56.888563');
$interval = $today->diff($mpsAgeT);
$diff = $interval->format('%y');
if (!($interval->d === 0 && $interval->m === 0 && $interval->h === 0 && $interval->i === 0 && $interval->s === 0)) {
    $diff++;
}
echo $diff;// returns 4

How to get year + 1 if the date is already passed

2 Answers2