php validate user is over 18

Question

Just wondering, can I do this to validate that a user has entered a date over 18?

//Validate for users over 18 only
function time($then, $min)
{
$then = strtotime('March 23, 1988');
//The age to be over, over +18
$min = strtotime('+18 years', $then);
echo $min;
if(time() < $min) 
{
die('Not 18'); 
}
}

Just stumbled across this function date_diff: http://www.php.net/manual/en/function.date-diff.php Looks, even more promising.

Answer 1

Why not? The only problem to me, is the User Interface - how you send out the error message elegantly to the user.

On another note, your function might not work properly as you did not intake a proper birthday (you are using a fixed birthday). You should change 'March 23, 1988' to $then

//Validate for users over 18 only
function time($then, $min)
{
  // $then will first be a string-date
  $then = strtotime($then);
  //The age to be over, over +18
  $min = strtotime('+18 years', $then);
  echo $min;
  if(time() < $min) 
  {
    die('Not 18'); 
  }
}

Or you can:

// validate birthday
function validate_age($birthday, $age = 18)
{

  // $birthday can be UNIX_TIMESTAMP or just a string-date.
  if(is_string($birthday)){
    $birthday = strtotime($birthday);
  }

  // check
  // 31536000 is the number of seconds in a 365 days year.
  if(time() - $birthday < $age * 31536000) 
  {
    return false;
  }

  return true;

}

Answer 2

if( strtotime("1988/03/23") < (time() - (18 * 60 * 60 * 24 * 365))) {
  print "yes";
} else {
  print "no";
}

...not accounting for leaps years however