Determine Age with PHP

aksival

Here's one for the mathematicians...

In an effort to determine a person's age (without using any function that utilizes the Unix timestamp) I have written the following script:

<?php
// Returns the user's age based on the current date
function currentAge($birthdate) {
    $current = date("Ymd");
    $birthdate = date("Ymd", strtotime($birthdate));
    $age = intval(($current - $birthdate) / 10000);
    return($age);
}

echo currentAge("09/28/1987"); // Hilary Duff's Birthday
?>

It appears remarkably accurate, but I know that the decimal value ($age before intval) will be slightly off. This is not important as all that is being determined is the age as an integer.

Can anyone find a reason / situation where this would ever fail? It appears to work even on leap years.

Best,
aksival

devinemke

aksival wrote:
...without using any function that utilizes the Unix timestamp...

...not true. you are using [man]strtotime[/man] whose whole job is converting strings to UNIX timestamps.

etully

This version is precise to the second and nothing will be "slightly off".

function currentAge($birthdate) { 
$age = floor((time()-strtotime($birthdate))/31536000);
return $age;
}

But the real question is this: You say that you want to determine whether the age is an integer. I assume by that, you mean that you want to know if today is their birthday? Of course, if I was born on November 7 and today is November 7 but it's 11:30 AM, then comparing the date is going to make me 11.5 hours off. So my age would be precisely 38 but because of the 11.5 hour discrepency, your function would determine that I am not precisely 38 years old so today must not be my birthday. Likewise, if today is the day before my birthday at 11:00 pm, then I am not quite 38 years old yet but your function would put me "close enough".

In my opinion, instead of checking to see if the age is an integer and rounding it... I would compare todays day / month against the day and month of the person's actual birthdate like this:

if (date("j/n")==date("j/n",strtotime($birthdate))) { print "today is their birthday"; }

And besides, what have you got against the Unix timestamp?!?!

aksival

I assume by that, you mean that you want to know if today is their birthday?

No, just trying to determine legal age. Given that, the time of birth is completely irrelevant. We're just considering the day, month and year here.

For example, if you were born in the evening on the 5th of June, 2000, then on the 5th of June, 2021 you will be 'legally' 21 years old, whether it's morning, noon or night. The algorithm seems to calculates the 'legal' age correctly.

And besides, what have you got against the Unix timestamp?!?!

It's a challenge? lol. I've nothing against it at all, actually 🙂 Just trying to see if the script can be proven incorrect mathematically, which I'm really starting to doubt.

aksival

...not true. you are using strtotime whose whole job is converting strings to UNIX timestamps.

I knew someone would catch that 🙂 And you are correct, however, strtotime() is used for convenience (so the user may enter 'June 5th, 2000' or '6/5/2000' or even 'Jun 5 2000' and it would still work).

I was actually referring to the mathematical operations not using the timestamp. Sorry about that 🙂