XML using Javascript in Google Chrome

Question

Hy all,

I am using a code that goes through an XML file and show me the data.. That code works great in IE, Firefox and Opera.. Now I would like to know how to configure it to work on Chrome..

Using the navigator.appName == "Microsoft Internet Explorer" and navigator.appName == "Netscape", I was capable of testing the identity of the browser used, and whether to use ActiveX objects or httpRequest.

Keeping in mind, when alerting the navigator.appName , on Chrome, Firefox and Opera, I get Netscape.

This is the full version of my code:

    <html>
    <body>
    <script type="text/javascript">

    alert(navigator.appName);
    if (navigator.appName == "Microsoft Internet Explorer") {
        xhttp = new ActiveXObject("Microsoft.XMLHTTP");
        xhttp.open("GET", "http://www.multimediaprof.com/test/emp2.xml", false);
    }
    else if (navigator.appName == "Netscape") {

        xhttp = new XMLHttpRequest();
        alert("step 1");

        xhttp.open("GET", "emp2.xml", false);
    }
    alert("step 2");
    xhttp.send("");
    alert("step 3");
    xmlDoc = xhttp.responseXML;
    alert("step 4");
    alert(xmlDoc);
    document.write(xmlDoc.documentElement.nodeName + " loaded");
    alert("step 5");
    var str = xmlDoc.getElementsByTagName("to")[0].childNodes[0].nodeValue;
    alert("step 6");
    alert(str);

</script>
</body>
</html>

New full version of code suggested now :

<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">

var xhttp, xmlDoc, str;

if (window.XMLHttpRequest) {
  xhttp = new XMLHttpRequest();
    xhttp.open("GET", "emp2.xml", false);
} else if (window.ActiveXObject) {
  xhttp = new ActiveXObject("Microsoft.XMLHTTP");
   xhttp.open("GET", "http://www.multimediaprof.com/test/emp2.xml", false);
} else {
  alert("Cannot create XmlHttpRequest object");
}
if (xhttp) {
  xhttp.send("");

  if (xhttp.responseXML != null) {
    xmlDoc = xhttp.responseXML;
    str = xmlDoc.getElementsByTagName("to")[0].childNodes[0].nodeValue;
    alert(str);
  } else {
    alert("Server response was invalid.");
  }
}
</script>
</head>
</html>

Tomalak · Accepted Answer · 2012-03-15 07:10:59Z

1

Just don't use appName at all.

var xhttp, xmlDoc, str;

if (window.XMLHttpRequest) {
  xhttp = new XMLHttpRequest();
} else if (window.ActiveXObject) {
  xhttp = new ActiveXObject("Microsoft.XMLHTTP");
} else {
  alert("Cannot create XmlHttpRequest object");
}
if (xhttp) {
  xhttp.open("GET", "emp2.xml", false);
  xhttp.send();

  if (xhttp.responseXML != null) {
    xmlDoc = responseXML;
    str = xmlDoc.getElementsByTagName("to")[0].childNodes[0].nodeValue;
    alert(str);
  } else {
    alert("Server response was invalid.");
  }
}

Note:

You must declare all the variables you use. There is no reason ever not to declare a variable.
You really should use asynchronous requests. Learn how to use them. Synchronous requests are bad and wrong.
You should never ever use document.write().
You should always check things before you use them. For example, you cannot simply use xhttp.responseXML without even checking that it exists.
The same goes for getElementsByTagName("to")[0].childNodes[0].nodeValue. This can fail at any time. If you never expect an error here, at least wrap it in a try/catch block and handle unexpected errors gracefully.
There is console.log(). That's a lot better than alert().

answered Mar 15, 2012 at 7:10

Tomalak

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6 Comments

ELias Rahme Over a year ago

Your code only worked on firefox and IE....and to make it work on them it needed tweeking

ELias Rahme Over a year ago

I still don't have access in google chrome...any ideas ? is there something special to announce or stuff?

Tomalak Over a year ago

The above works for me in Google Chrome, no change or tweaking required. You are doing something else wrong.

ELias Rahme Over a year ago

In chrome i get : Uncaught Error : NETWORL_ERR: XMLHttpRequest Exception 101 right under the xhttp.send("");

ELias Rahme Over a year ago

I solved the problem and it runs smoothly on firefox, opera and IE but still that same error on chrome, and that error i already had it with my own basic code...so that's no new error! please help anyone

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