Alder i dag

<?php
$q = mysql_query("SELECT id, name, flags, DATE_FORMAT(date, '%d') AS dato, DATE_FORMAT(date, '%m') AS maaned, DATE_FORMAT(date, '%Y') AS aar FROM turen WHERE MONTH(date) = '".date('n')."' AND DAYOFMONTH(date) = '".date('j')."' ORDER BY name ASC") or die(mysql_error());
while($vis = mysql_fetch_array($q)){
$id = $vis[id]; // Brug quotes om dine indekses - altså $vis["id"] og ikke $vis[id]
$name = $vis[name];
$flags = $vis[flags];
$dato = $vis[dato];
$maaned = $vis[maaned];
$aar = $vis[aar];
$date = $dato-$maaned-$aar; // her trækker du tal fra hinanden

$idag = date("Y") . date("m") . date("d");
$date = $date; // her siger du bare at du vil sætte variablen $date til variablen $date.
$alder = substr($idag - $date,0,-4);

echo "$alder <img src=$flags width='15' height='10' border=0>  <a href=http://www.tourfacts.dk/profilryttere.php?id=$id class='link2'>$name</a><br>";

}

?>

<?php
  $var1 = "hej ";
  $var2 = "med dig";
  $streng = $var1 . $var2; // bliver til "hej med dig";
?>

<?php
...
  $dato = $vis["dato"];
  $maaned = $vis["maaned"];
  $aar = $vis["aar"];
  
  $alder = date("Y") - $aar;
  if( mktime(0, 0, 0, $maaned, $dato, date("Y")) < time() )
    $alder = $alder - 1;
...
?>