Coaches
< Back
$i = "1";
$ipag = "0";
$resultaat=mysql_query("SELECT * FROM coaches WHERE actief = '1' ORDER BY Naam") or die(mysql_error('Er is iets fout gegaan met de database'));
while($rij=mysql_fetch_object($resultaat)){
if ($i%6==1) { $ipag=$ipag+1; echo "$ipag";}
$i++;
}
?>
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