* Thus wrote John Ryan ([EMAIL PROTECTED]):
> Hi,
>
hello ryan,
> In mySQL, I store dates as YYYY-MM-DD, a standard DATE type. It stores users
> date of births. I need to calculate in a PHP script, the users age from this
> DOB. I get a PHP date in the same format as the mySQL and subtract, which
> returns the year rounded off. ie, it doesnt matter if your birthdays in june
> of 1983 and the date is januray 2003, your age is still returned as 20, when
> it should be 19.
>
> Does anyone know how can i get the right age?
To get a real age of someone you would use seconds, but since the actual
count of seconds get messed up on leap years (through the convertion
from seconds to years), I approached it with the method most people
calculate ages.
Most people first look at the year and find the differnce, then if
the month/day hasn't come around you subtract one; alas the common
calculation for age.
Here is the code to demonstrate that logic:
$dob = split('-', '1980-12-1');
$now = split('-', date('Y-m-d'));
// we are either this age or one less
$age = $now[0] - $dob[0];
// have we gotten to the month of in dob?
if ($now[1] < $dob[1]) {
// no, so we are technically a year less.
$age--;
// If we're in the month, has the day come yet?
} elseif ($now[1] = $dob[1] && $now[2] < $now[3]) {
// no, still a few more days.
$age--;
}
// your age is proper, in day to day usage.
/*
Note:
instead of the if () elseif() you can use
if (mktime(0,0,0,$now[1],$now[2]) < mktime(0,0,0,$dob[1],$dob[2])) {
$age--;
}
*/
HTH,
Curt
--
"I used to think I was indecisive, but now I'm not so sure."
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