Hi....

I have form which I put save function on the last textbox:
here is my code:

Code:
<?php
    error_reporting(0);
   date_default_timezone_set("Asia/Singapore"); //set the time zone  
$con = mysql_connect('localhost', 'root','');

if (!$con) {
    echo 'failed';
    die();
}
mysql_select_db("mes", $con);
?>

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> 
<html> 
<head> 
<script type="text/javascript" >  

var input_size = 1;

function checkTextBox(bc){
   var barcode_ =  bc.tabIndex;
   
   if ( bc.value.length > input_size ) 
        { 
            for(i=0; i<document.barcode.elements.length; i++) 
            { 
                      if( document.barcode.elements[i].tabIndex == (barcode_+1) ) 
                     {    
                         document.barcode.elements[i].focus(); 
                         break; 
                     } 
          } 
    }         
} 

function postSet() {
    if (window.event.keyCode==13 || window.event.keyCode==10) {
        document.getElementById('code_read_box6').disabled = true;
        save();
        alert('code_read_box6');
    }
}
 
</script> 

<script type="text/javascript"> 
var ajaxTimeOut = null;
var ajaxTimeOutOperator = null; 
var responsePHP; // = "no_reply"
var responsePHPOperator;
var changeFocus; //= false;
var transactionWasSaved;

function remoteRequestObject() {
    var ajaxRequest = false;
    try {
        ajaxRequest = new XMLHttpRequest();
    }
    catch(err) {
        try{
            ajaxRequest = new ActiveXObject("MSxml2.XMLHTTP");
        }
        catch(err) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            }
            catch(err){
                // --> change to DOM alert("Not Supported Browser") + err.description;
                notify('Not Supported Browser.');
                return false;
            }
        }
    }
    return ajaxRequest;
} 

var ajaxRequest; // = remoteRequestObject();
var ajaxRequestOperator;
</script>

<script type="text/javascript">
function save() {            
    ajaxRequest.onreadystatechange = function () {
    if (ajaxRequest.readyState==4 && ajaxRequest.status==200) {
       var result = ajaxRequest.responseText;
       
        alert (result);
           
           if (result == "failed") {
            document.getElementById('code_read_box6').disabled = false;
            document.getElementById('code_read_box6').value = "";
            document.getElementById('code_read_box6').focus();
            notify("Please scan again.");
           }
           
           if (result == "saved") {
              alert(result);
              notify("Transaction has been saved.");
              reset();
           }   
           
       }     
    }      
   
    
 var url = "save_barcode.php";   

ajaxRequest.open("POST", url, true);
ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
ajaxRequest.setRequestHeader("Content-length", parameters.length);
ajaxRequest.setRequestHeader("Connection", "close");
ajaxRequest.send(parameters);
}
 
</script> 
 
</head> 
<body onLoad="document.barcode.code_read_box1.focus();"> 
<form name="barcode" > 
<input type="text" tabindex="1" id="code_read_box1" value="" onkeyup="checkTextBox(this);"/><br/> 
<input type="text" tabindex="2" id="code_read_box2" value="" onkeyup="checkTextBox(this);"/><br/> 
<input type="text" tabindex="3" id="code_read_box3" value="" onkeyup="checkTextBox(this);"/><br/> 
<input type="text" tabindex="4" id="code_read_box4" value="" onkeyup="checkTextBox(this);"/><br/> 
<input type="text" tabindex="5" id="code_read_box5" value="" onkeyup="checkTextBox(this);"/><br/> 
<input type="text" tabindex="6" id="code_read_box6" value="" onkeyup="checkTextBox(this);" onkeypress="postSet()"/><br/> 
</form> 
</body> 
 
</html>
I got an error:
'ajaxRequest' is null or not an object on line 72

It display the error when I press enter on the last textbox.

Sorry, I'm not familiar in ajax..I hope somebody can help me

Thank you