I had a customer drop down if i select customer i should fetch the addaress and tinno

[Praveen040]

Please check this code and reply me .....

<script language="JavaScript">
 
    var listObj=document.getElementById('customer');
 
    listObj.selectedIndex=0;
//    var ind=document.getElementById('list').selectedIndex;
//alert(ind);
 
    if(listObj.value!=""){
        showUser(listObj.value)     
    }
        document.getElementById('customer').style.width="158";
 
//    document.getElementById('client_list').selectedIndex=0;
    </script>
 
 
 
var xmlHttp;
 
var isIE = false;
 
// on !IE we only have to initialize it once
if (window.XMLHttpRequest) {
    xmlHttp = new XMLHttpRequest();
}
 
function showUser(str)    
{
 
if (window.XMLHttpRequest) {
       // branch for IE/Windows ActiveX version
 
   } else if (window.ActiveXObject) {
       xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
   }
 
   if (xmlHttp==null)
   {
    alert ("Browser does not support HTTP Request");
    return;
  }
 
 
 var url="php/invoicedisply.php"
 
 url=url+"?q="+str
 url=url+"&frm=Client"
 url=url+"&sid="+Math.random()
 
 xmlHttp.onreadystatechange=processChanged
 xmlHttp.open("GET",url,true)
 xmlHttp.send(null)
 }
 
 function processChanged()
{
 
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{
 
 
 xmlDoc=xmlHttp.responseXML;
 
document.getElementById("address1").value=xmlDoc.getElementsByTagName("address1")[0].childNodes[0].nodeValue;
document.getElementById("tinno").value=xmlDoc.getElementsByTagName("tinno")[0].childNodes[0].nodeValue;
}
}
 
 
 
 
<?php
    header('Content-Type: text/xml');
    header("Cache-Control: no-cache, must-revalidate");
 
    require("db_connect.php");
 
    $frm=$_GET['frm'];
    if($frm=="Client")
    {
        $q=$_GET["q"];
 
       $sql="SELECT address1,tin_no FROM customermaster WHERE customer_id='".$q."'";
 
        //echo $sql;
        $result = mysql_query($sql,$link);
        checkError($result);
        echo '<?xml version="1.0" encoding="ISO-8859-1"?><customer>';
 
        while($row = mysql_fetch_array($result))
    {
 
           echo "<address1>". $row['address1']."</address1>";
 
             echo "<tinno>". $row['tin_no']."</tinno>";
 
        }
        echo "</customer>";
    }
 
 
 
    mysql_close($link);
    die();
 
    function checkError($result)
    {
        if(!$result){
            echo "<font size=\"4\">MySQL Error : ".mysql_error()."</font>";
            mysql_error();
            die();
        }
    }
 
?>

[Dormilich]

and what exactly is the problem?