[Resolved] How old?

pablodelapena

Ive searched the forums and google and most places but ic ant find any help!

I want to be able to calculate how old someone is (in years only) from their DOB.

the dob is stored as 3 seperate values in MySQL

i.e day, month, and year all in number format (i.e 01 12 1986)

this is proving to be difficult!

any help appreciated!

Pablo de la Pena

parawizard

well firstly ur gonna have to ether pull out the years out of the date ( i dont know how to do this )

then ur gonna have to compare it with the date now

// u bring in ur DOB and take out the years and put them in $DOB

$Date = date("Y");

$Difference=$Date-$DOB;

echo $DOB;    //to see how old the person is

hope that helps somewhat maybe someone else on the forum would kindly give u the info on spliting the years out of DOB

planetsim

Its not pretty but it works

$date['month'] = 11; 
$date['day'] = 1; 
$date['year'] = 1986;

$bday = mktime(0,0,0,$date['month'],$date['day'],$date['year']);
echo floor(((time()-$bday) / 60 / 60 / 24 / 365));

Weedpacket

So I guess that someone who was born on July 20, 1969 is thirty-three in 2003 up until July 20, 2003, at which point they turn 34?

list($tyear,$tmonth,$tday)=explode(' ',date('Y m d'));
$byear=1969;
$bmonth=7;
$bday=20;
$age=$tyear-$byear;
if($tmonth<$bmonth || ($tmonth==$bmonth && $tday<$bday))
$age--;
echo $age;

pablodelapena

awesome Weedpacket! yours worked a treat!

Planetsim. thanks anyway but your didnt seem to work for me. when i put an age in (mine, im 17) it told me i was 3 years old. summin wrong there! thanks all the same!

PHPbuilder boards once again save the day!