PHP calculate age

Question

I'm looking for a way to calculate the age of a person, given their DOB in the format dd/mm/yyyy.

I was using the following function which worked fine for several months until some kind of glitch caused the while loop to never end and grind the entire site to a halt. Since there are almost 100,000 DOBs going through this function several times a day, it's hard to pin down what was causing this.

Does anyone have a more reliable way of calculating the age?

//replace / with - so strtotime works
$dob = strtotime(str_replace("/","-",$birthdayDate));       
$tdate = time();

$age = 0;
while( $tdate > $dob = strtotime('+1 year', $dob))
{
    ++$age;
}
return $age;

EDIT: this function seems to work OK some of the time, but returns "40" for a DOB of 14/09/1986

return floor((time() - strtotime($birthdayDate))/31556926);

Answer 1

i18n :

function getAge($birthdate, $pattern = 'eu')
{
    $patterns = array(
        'eu'    => 'd/m/Y',
        'mysql' => 'Y-m-d',
        'us'    => 'm/d/Y',
    );

    $now      = new DateTime();
    $in       = DateTime::createFromFormat($patterns[$pattern], $birthdate);
    $interval = $now->diff($in);
    return $interval->y;
}

// Usage
echo getAge('05/29/1984', 'us');
// return 28

Answer 2

I have found this script reliable. It takes the date format as YYYY-mm-dd, but it could be modified for other formats pretty easily.

/*
* Get age from dob
* @param        dob      string       The dob to validate in mysql format (yyyy-mm-dd)
* @return            integer      The age in years as of the current date
*/
function getAge($dob) {
    //calculate years of age (input string: YYYY-MM-DD)
    list($year, $month, $day) = explode("-", $dob);

    $year_diff  = date("Y") - $year;
    $month_diff = date("m") - $month;
    $day_diff   = date("d") - $day;

    if ($day_diff < 0 || $month_diff < 0)
        $year_diff--;

    return $year_diff;
}

Answer 3

  function dob ($birthday){
    list($day,$month,$year) = explode("/",$birthday);
    $year_diff  = date("Y") - $year;
    $month_diff = date("m") - $month;
    $day_diff   = date("d") - $day;
    if ($day_diff < 0 || $month_diff < 0)
      $year_diff--;
    return $year_diff;
  }

Answer 4

Try this :

<?php
  $birth_date = strtotime("1988-03-22");
  $now = time();
  $age = $now-$birth_date;
  $a = $age/60/60/24/365.25;
  echo floor($a);
?>

Answer 5

Simple method for calculating Age from dob:

$_age = floor((time() - strtotime('1986-09-16')) / 31556926);

31556926 is the number of seconds in a year.

Answer 6

If you want to only get fullyears as age, there is a supersimple way on doing that. treat dates formatted as 'YYYYMMDD' as numbers and substract them. After that cancel out the MMDD part by dividing the result with 10000 and floor it down. Simple and never fails, even takes to account leapyears and your current server time ;)

Since birthays or mostly provided by full dates on birth location and they are relevant to CURRENT LOCAL TIME (where the age check is actually done).

$now = date['Ymd'];
$birthday = '19780917'; #september 17th, 1978
$age = floor(($now-$birtday)/10000);

so if you want to check if someone is 18 or 21 or below 100 on your timezone (nevermind the origin timezone) by birthday, this is my way to do this