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newphpbees
03 May 2012, 03:46 AM
Hi....

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