Setting a users age

[graham23s]

Hi Guys,

 

when i get the users birthyear from mysql i do a simple bit of math to get the users age:

 

<?php
// work out the users age //
$year = date('Y');
$users_age = $year - $users_birth_year;
?>

 

$users_birth_year contains 1979 for example so

 

2008 - 1979 = 28

 

but im trying to think of the best way to get exactly the users age, so when i turn 29 in june how to update the age then?

 

thanks guys

 

Graham

[peytonrm]

It's not possible to do that with only the year. You would need to store a full date/time in the database and then get a full timestamp (e.g. date('Y-m-d H:i:s') in your code and subtract those two. 

[graham23s]

sorry i should have said i also store the users birthmonth and day to e.g

 

18 | 6 | 1979

 

Graham

 

 

[PFMaBiSmAd]

First of all, you should be storing dates in a DATE type field. Then you can find the age using a simple formula in a query - http://dev.mysql.com/doc/refman/5.0/en/date-calculations.html

[graham23s]

Hi Mate,

 

ah i see, the way i have set it i have:

 

<?php
print("<td class='form_style' align='left'><label><b>Birthdate:</b></label></td><td class='form_style' align='left'>\n");
print("<select name='birth_month'>\n");
foreach($birth_month as $key => $birth_month_value)
{
  print("<option value='$key'>$birth_month_value</option>\n");
}
print("</select>\n");

print("<select name='birth_day'>\n");
for($i = 1; $i <= 31; $i++)
{
  print("<option value='$i'>$i</option>\n");
}
print("</select>\n");

print("<select name='birth_year'>\n");
for($year = 1930; $year <= 2008; $year++)
{
  print("<option value='$year'>$year</option>\n");
}
print("</select>\n");
print("</td>\n");
?>

 

i have been individually storing the month,day,year into mysql seperately, instead of all in the 1 column

 

whats the best way to do all 3 in the 1 column? just grab all 3 values then stitch them together before the mysql insertion:

 

$dob = $bmonth . "-" . $bday . "-" .  $byear;

 

or is there a better way?

 

thanks mate

 

Graham

[Northern Flame]

heres a function i created that calculates the age

of my users.

 

$birth = mm/dd/yyyy

ex: $birth = 01/04/1967

 

heres the function:

<?php
	function age($birth){
		$blown = explode("/", $birth);

		$year = date("Y");
		$month = date("m");
		$day = date("d");

		$n_year = $year - $blown[2];
		$n_month = $month - $blown[1];
		$n_day = $day - $blown[0];

		if($n_month < 0){
			$age = $n_year - 1;
		} else{
			if($n_day < 0){
				$age = $n_year - 1;
			} else{
				$age = $n_year;
			}
		}
		return $age;
	}

$birth = "01/02/1967";

age($birth);
?>

[quiettech]

Check the php manual on DateTime functions. Particularly the date_create() function which will create a DateTime object for you, you can then use for date/time arithmetics or for SQL insertion.

 

A typical example would probably be:

 

$dateobj = datecreate('08 23 1969');

 

You can test $dateobj for 'false' in case the data is invalid.

 

EDIT:

Alternatively you can also use:

 

$timestamp = strtotime('08 23 1969');

 

It has the same effect and probably the best option for SQL insertion. However probably not cross-compatible with other databases not being served on UNIX.

 

EDIT2: both functions use the British form of month, day, year. Corrected my own examples.

[AndyB]

Whatever else you do, you can avoid lots of future pain by storing dates in an unequivocal format. Using the ISO yyyy-mm-dd for stored date format allows you flexibility and simplicity that will never exist if you choose some quaint local format for 'date'.  Note that storing date in a rational format does not preclude displaying it in whatever weird form floats your boat.