อยากได้วิธีเช็ค username ซ้ำ อ่ะคับ แล้วขึ้น alert message ทันที ใครพอทราบบ้างคับ

<input type="text" id="NRIC" /><span id="sNRIC"></span>
<input type="email" id="Email" /><span id="sEmail"></span>

<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
$(function(){
	
	$("#NRIC,#Email").change(function(){
		$("#sNRIC").empty();
		$("#sEmail").empty();
		$.ajax({url:"returnCustomer1.php",
			async:false,
			global:false,
			cache:false,
			dataType:"post",
			data:'NRIC=' +$("#NRIC").val()+
				'&Email='+$("#Email").val(),
			success: function(result){
				var obj = jQuery.parseJSON(result);
				if(obj != ''){
					$.each(obj,function(key,val){
						if($('#NRIC').val() == inval['NRIC']){
							$("#sNRIC").html(" <font color='red'>รหัสลูกค้านี้มีอยู่แล้ว</font>");
						}
						if($('#Email').val() == inval['Email']){
							$("#sEmail").html(" <font color='red'>รหัสลูกค้านี้มีอยู่แล้ว</font>");
						}
					});
				}	
			}
		});
	});
});

<?
$objConnect = mysql_connect("localhost","root","") or die(mysql_error());
$objDB = mysql_select_db("uog");
$strSQL = "SELECT * FROM member WHERE 1 AND NRIC = '".$_POST["NRIC"]."' OR Email = '".$_POST["EMail"]."' ";
$objQuery = mysql_query($strSQL) or die (mysql_error());
$intNumField = mysql_num_fields($objQuery);
$resultArray = array();
while($obResult = mysql_fetch_array($objQuery))
{
$arrCol = array();
for($i=0;$i<$intNumField;$i++)
{
$arrCol[mysql_field_name($objQuery,$i)] = $obResult[$i];
}
array_push($resultArray,$arrCol);
}

mysql_close($objConnect);

echo json_encode($resultArray);
?>