Age calculation

RdP

I have a script running on my site. Sort of profile type kinda site. It calculates the age of people by looking at their Birthday, Birthmonth and Birthyear and the current date. However, there is a problem: this way of calculation doesn't display the correct age for all people.

$stampmebaby =getdate(time()) ;
$day= $stampmebaby[mday] ;
$month=$stampmebaby[mon] ;
$year=$stampmebaby[year] ;
$age_year= $year-$row[byear]   ;
$age=$age_year     ;

if ($month>=$row[bmonth] AND $day>=$row[bday])
$age=$age_year      ;
else
$age=$age_year-1 ;

I have tried to explain to myself exactly how the calculation is done in normal, human language... This is what I came up with:

Age = Year - Birthyear
If month is bigger than birthmonth AND day is bigger than birthday
Age = Year - Birthdate
otherwise
Age = Year - Birthdate - 1

Now this would work today (november 21st 2002) for someone who's born on September 17th 1985, because 21 is more than 17. But for someone who's born on January 29th 1986 it won't work, because 29 is more than 21 and then it will jump to the "else $age=$age_year -1" part. This part is necessary sometimes to calculate the correct age.. but not in this case.

I know this might sound complicated.. but please, can anyone give me suggestions on what the proper code would be if I wanted to calculate someone's age in PHP when I was given their birthday, birthmonth and birthyear?

largo

Rdp,

This may be a confusing way of going about it. I would recommend using mktime to generate timestamps and work with those.

As an example to start off with:

<?php
// required vars.
$stampmebaby =getdate(time());
$day= $stampmebaby[mday];
$month=$stampmebaby[mon];
$year=$stampmebaby[year];

// say user input results in
$mday = 7;
$mmon = 04;
$myear = 1977;

// generate the time-stamps
$now = mktime(0, 0, 0, $month, $day, $year);
$then = mktime(0, 0, 0, $mmon, $mday, $myear);

// convert time-stamp to years.
$userage = ($now - $then) / 31536000;
echo $userage;
?>

you'll probably get a xx.xxxxxx kinda result (in years). You can get creative with that or use a better way to decode the timestamp.

-m.

pnoeric

Hey, I adapted your code but I must be missing something-- I'm trying to compute if someone is 18 or over, and the code keeps telling me they're under 18 when they're not:


// get birthday
$m = $_REQUEST['r_bmonth'];
$d = $_REQUEST['r_bday'];
$y = $_REQUEST['r_byear'];
if ($y < 1900) $y = $y + 1900;		// correct 2-digit years

// generate the time-stamps
$now = time();  // mktime(0, 0, 0, $month, $day, $year);
$then = mktime(0, 0, 0, $m, $d, $y);

// convert time-stamp to years.
$userage = ($now - $then) / 31536000;

echo "Age: $userage";

If I run this code on the birthday 7/26/84, I get 17.047 years old, which is wrong-- the person turned 18 last year, 7/26/2002! The date on my server is correct. What's going on?

Hate these dumb/obvious things ;-)

best
Eric

xblue

I don't know where this huge difference comes from, but you'd probably be off by some days in most cases since you disregard leap years.

For more precision, I'd add 18 to the birth year and compare it to the current date. A numeric comparision should do if we have the data in YYYYMMDD format. Example:

$day = $_GET['day']; // make sure you receive it it in 2 digit form 
$month = $_GET['month']; // make sure you receive it it in 2 digit form 
$year = $_GET['year']; // make sure your receive it in 4 digit form

$today = date('Ymd');
echo $today."<br />";
$birthday18 = ($year+18).$month.$day;
echo $birthday18."<br />";

// casting probably is not even needed
if (intval($today) >= intval($birthday18)) {
   echo "old enough";
} else {
   echo "too young";
}

pnoeric

Oh, duh... talk about over-thinking the problem ;-) Thanks for this most-obvious solution. best E