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Home/ Questions/Q 266353
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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T22:50:53+00:00 2026-05-11T22:50:53+00:00

I’m looking for a function that calculates years from a date in format: 0000-00-00.

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I’m looking for a function that calculates years from a date in format: 0000-00-00.
Found this function, but it wont work.

// Calculate the age from a given birth date
// Example: GetAge("1986-06-18");
function getAge($Birthdate)
{
  // Explode the date into meaningful variables
  list($BirthYear,$BirthMonth,$BirthDay) = explode("-", $Birthdate);
  // Find the differences
  $YearDiff = date("Y") - $BirthYear;
  $MonthDiff = date("m") - $BirthMonth;
  $DayDiff = date("d") - $BirthDay;
  // If the birthday has not occured this year
  if ($DayDiff < 0 || $MonthDiff < 0)
  $YearDiff--;
 }

echo getAge('1990-04-04');

outputs nothing :/
i have error reporting on but i dont get any errors

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  1. Editorial Team
    Editorial Team
    2026-05-11T22:50:53+00:00Added an answer on May 11, 2026 at 10:50 pm

    Your code doesn’t work because the function is not returning anything to print.

    As far as algorithms go, how about this:

    function getAge($then) {
        $then_ts = strtotime($then);
        $then_year = date('Y', $then_ts);
        $age = date('Y') - $then_year;
        if(strtotime('+' . $age . ' years', $then_ts) > time()) $age--;
        return $age;
    }
    print getAge('1990-04-04'); // 19
    print getAge('1990-08-04'); // 18, birthday hasn't happened yet
    

    This is the same algorithm (just in PHP) as the accepted answer in this question.

    A shorter way of doing it:

    function getAge($then) {
        $then = date('Ymd', strtotime($then));
        $diff = date('Ymd') - $then;
        return substr($diff, 0, -4);
    }
    
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