Find out current age

Hello,

I have birth dates entered in my database entered in the form of numerals. So someone born on January 20th, 1990 would be 1 20 1990. I am trying to create a script so it will tell you your current age. I tried just using the current year, 2009 subtracted by there year of birth, but it is not accurate. I have to use the months and days as well, not just the year. Does anyone know how to do this?

Thank You,
Jason

 

< ?php $dob = mktime(0, 0, 0, 10, 10, 1950); echo round(((time() - $dob) / 31557600), 0); ?>

This is the code I use on a page of mine to display my age in years - if you exchange the mktime() with the values from your database, it should work (you might have to pull it with some type of formatting, but it should work).

Hope it gets you started, at least.

(And no, that's not my birthdate )

 

PHP won't know its January if you strtotime like this:
$time = strtotime("1 20 1990");

So, (might not be the best way but it works)

$dob = explode(" ","1 20 1990");
$dob_stamp = mktime(0, 0, 0, $dob[0], $dob[1], $dob[2]);
$one_year_in_secs = 60*60*24*365;
$age = floor($dob_stamp/$one_year_in_secs);
echo $age;

PHP:

I use floor() instead of round()

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I forgot to say that I have the day, month and year in 3 different fields, they are not all together. How do I edit the script to make it for like that?

 

After explode() using blankspace " ",
- $dob[0] carries the month
- $dob[1] carries the day
- $dob[2] carries the year

So,

$dob_stamp = mktime(0, 0, 0, $yoursql_month, $yoursql_day, $yoursql_year);
$one_year_in_secs = 60*60*24*365;
$age = floor($dob_stamp/$one_year_in_secs);
echo $age;

PHP:

Just replace $yoursql_month, $yoursql_day, $yoursql_year with the result from sql.

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