Hey,
finally solved this, incase anyones interested heres how i did it :
$day1=strtotime($age); // eg: 1979-12-07
$day2 = strtotime(date("Y-m-d"));
$dif_s = ($day2-$day1);
$dif_d = ($dif_s/60/60/24);
$dif_y = round(($dif_d/365.24));
echo "The person is $dif_y year(s) old";
Cheers,
Ryan
----------------------------------
Hey,
Anybody have an age function where i can pass a date object and get back an
int?
eg:
$years_old=get_years("1979-12-07");
Thanks,
Ryan
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