Calculating age from date of birth in PHP - php

Question

What's the most precise function you have come across to work out an age from the users date of birth. I have the following code and was wondering how it could be improved as it doesn't support all date formats and not sure if it's the most accurate function either (DateTime compliance would be nice).
function getAge($birthday) {
return floor((strtotime(date('d-m-Y')) - strtotime($date))/(60*60*24*365.2421896));
}

Answer 1

$birthday = new DateTime($birthday);
$interval = $birthday->diff(new DateTime);
echo $interval->y;
Should work

Answer 2

Check this
<?php
$c= date('Y');
$y= date('Y',strtotime('1988-12-29'));
echo $c-$y;
?>

Answer 3

Use this code to have full age including years, months and days-
<?php
//full age calulator
$bday = new DateTime('02.08.1991');//dd.mm.yyyy
$today = new DateTime('00:00:00'); // Current date
$diff = $today->diff($bday);
printf('%d years, %d month, %d days', $diff->y, $diff->m, $diff->d);
?>

Answer 4

Try using DateTime for this:
$now = new DateTime();
$birthday = new DateTime('1973-04-18 09:48:00');
echo $now->diff($birthday)->format('%y years'); // 49 years
See it in action

Answer 5

This works:
<?
$date = date_create('1984-10-26');
$interval = $date->diff(new DateTime);
echo $interval->y;
?>
If you tell me in what format your $birthday variable comes I will give you exact solution

Answer 6

WTF?
strtotime(date('d-m-Y'))
So you generate a date string from the current timestamp, then convert the date string back into a timestamp?
BTW, one of the reasons it's not working is that strtotime() assumes numeric dates to be in the format m/d/y (i.e. the US format of date first). Another reason is that the parameter ($birthday) is not used in the formula.

Answer 7

Change the $date to $birthday.

Answer 8

For supper accuracy you need to account for the leap year factor:
function get_age($dob_day,$dob_month,$dob_year){
$year = gmdate('Y');
$month = gmdate('m');
$day = gmdate('d');
//seconds in a day = 86400
$days_in_between = (mktime(0,0,0,$month,$day,$year) - mktime(0,0,0,$dob_month,$dob_day,$dob_year))/86400;
$age_float = $days_in_between / 365.242199; // Account for leap year
$age = (int)($age_float); // Remove decimal places without rounding up once number is + .5
return $age;
}
So use:
echo get_date(31,01,1985);
or whatever...
N.B. To see your EXACT age to the decimal
return $age_float
instead.

Answer 9

This function works fine.
function age($birthday){
list($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--;}
if ($day_diff < 0 && $month_diff < 0){$year_diff--;}
return $year_diff;
}
See BLOG Post

Answer 10

Here is my long/detailed version (you can make it shorter if you want):
$timestamp_birthdate = mktime(9, 0, 0, $birthdate_month, $birthdate_day, $birthdate_year);
$timestamp_now = time();
$difference_seconds = $timestamp_now-$timestamp_birthdate;
$difference_minutes = $difference_seconds/60;
$difference_hours = $difference_minutes/60;
$difference_days = $difference_hours/24;
$difference_years = $difference_days/365;