[SOLVED] Work out age from date of birth.

[DeanWhitehouse]

I dunno where to start with this,

 

i need to work out the age of a person from there date of birth.

the date is wrote like.

daymonthyear

28081991

 

any ideas?

[DarkWater]

You just need how many years old they are right, not like...days?

[DeanWhitehouse]

erm, yer how many years, but i also need to work out that they are say.

 

e.g.

 

bday = 28/08/91

 

then they will be 16 now not 17 but after then they will be 17 .

 

If that makes sense.

[rajivgonsalves]

You can try

 

<?php
$strDate = "28081991";
echo floor((time()-strtotime(substr($strDate,4,4)."-".substr($strDate,2,2)."-".substr($strDate,0,2)))/(86400*365));
?>

[Barand]

see

http://www.phpfreaks.com/forums/index.php/topic,180194.msg803782.html#msg803782

[DeanWhitehouse]

@rajivgonsalves

I used your code

$day = trim(htmlentities($_POST['age']));
$month =trim(htmlentities($_POST['age1']));
$year =trim(htmlentities($_POST['age2']));
$dob = $day.$month.$year;

//Work out birthday
$strDate = $dob;
echo floor((time()-strtotime(substr($strDate,4,4)."-".substr($strDate,2,2)."-".substr($strDate,0,2)))/(86400*365));
//end bday

 

but this works for my birthday and probably more, but when i submitted a form with the birthday being

01011920

it got the age 38.

 

any ideas?

[DeanWhitehouse]

Ok, i solved it, i need to have it so it's like

01011920

instead of

1011920

[corbin]

You just need years, right?

 

Different solution of the one earlier presented.  In my opinion, this one is cleaner.  Not sure if it's more efficient.

 

<?php

$month = 2;
$day = 18;
$year = 1992;

$tyear = (int) date('Y');
$tmonth = (int) date('n');
$tday = (int) date('j');
$years = $tyear - $year - 1;
if($tmonth > $month || ($tmonth == $month && $tday >= $day)) {
//oddly written, but i think that's the most effecient way to do it.
++$years;
}
//if the current date is past (or equal to) the birthday, add a year

echo $years;

?>

 

 

Edit:  Bleh, just realized the "if($month <= date('n') && $day <= date('j')) ++$years;" part is wrong.  Since you already have a solution, I'm not gonna bother fixing it, unless you're interested in this.

 

 

Edit:  Fixed it, because it bothered me not having it correct. lol