Simple XMLHttpRequest (Google Weather)

Question

Hello I want to get xml from Google Weather

var xmlhttp;

if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp= new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }

 xmlhttp.open("GET", "http://www.google.com/ig/api?weather=london&hl=en", true);

xmlhttp.send(null);

xmlDoc=xmlhttp.responseXML;

It`s not working . Thanks

We see this "it's not working" problem a lot. That's just not descriptive enough. **What** is not working? **How** is it failing? — Matt Ball, Jun 20 '11 at 15:48

score 3 · Answer 1 · edited Jun 20 '20 at 09:12

XMLHttpRequest is asynchronous. You need to use a callback. If you don't want to use a full-fledged library, I recommend using Quirksmode's XHR wrapper:

function callback(xhr)
{
    xmlDoc = xhr.responseXML;
    // further XML processing here
}

sendRequest('http://www.google.com/ig/api?weather=london&hl=en', callback);

If you absolutely insist on implementing this yourself:

// callback is the same as above

var xmlhttp;

if (window.XMLHttpRequest)
{
    xmlhttp = new XMLHttpRequest();
}
else
{
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}

xmlhttp.open("GET", "http://www.google.com/ig/api?weather=london&hl=en", true);

xmlhttp.onreadystatechange = function ()
{
    if (xmlhttp.readyState != 4) return;
    if (xmlhttp.status != 200 && xmlhttp.status != 304) return;
    callback(xmlhttp);
};

xmlhttp.send(null);

Edit

As @remi commented:

I think you'll get a cross domain access exception : you can't make an ajax request to an other domain than your page's. no ?

Which is (for the most part) correct. You'll need to use a server-side proxy, or whatever API that Google provides, instead of a regular XHR.

hmm I think you'll get a cross domain access exception : you can't make an ajax request to an other domain than your page's. no ? — remi bourgarel, Jun 20 '11 at 16:13
Ah, yes, I was so distracted by the OP's code that I didn't even look at the URL. — Matt Ball, Jun 20 '11 at 16:22

score -1 · Answer 2 · answered Jun 20 '11 at 17:15

-1

You can't do this via javascript to to it being a cross-domain request. You'd have to do this server-side.

In PHP you'd use CURL.

What you are trying to do can't be done with Javascript.

answered Jun 20 '11 at 17:15

timw4mail

1,716
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score -1 · Answer 3 · answered Jun 21 '11 at 07:06

Ok here is the code :

<html>
<body>

<script type="text/javascript">

var xmlhttp;
var xmlDoc;
function callback(xhr)
{
    xmlDoc = xhr.responseXML;
    // further XML processing here
}


if (window.XMLHttpRequest)
{
    xmlhttp = new XMLHttpRequest();
}
else
{
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}

xmlhttp.open("GET", "http://www.google.com/ig/api?weather=london&hl=en", true);
xmlhttp.onreadystatechange = function ()
{
    if (xmlhttp.readyState != 4) return;
    if (xmlhttp.status != 200 && xmlhttp.status != 304) return;
    callback(xmlhttp);
};

xmlhttp.send(null);



alert(xmlDoc);

</script>

</body>
</html>

It doesn`t returns any errors but alert returns undefined.

As I said in my answer, the execution is **asynchronous.** You need to move the alert into `callback`. — Matt Ball, Jun 21 '11 at 12:43

Simple XMLHttpRequest (Google Weather)

3 Answers3

Edit