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problem fetching remote xml file using ajax

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rammi07

Programmer
Aug 1, 2006
20
NL
Hi all i am using the follwoing code to display a feed but when run i get the following error:access is denied

xmlobj.open('GET',doc,true); ===>access is denied. error line 34

If i download the same xml file manually and place it in same folder as script then i do not get error. could any one tell me how to fix this problem so all visitor be able to use this script.Thanks



Code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"[URL unfurl="true"]http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">[/URL]
<html>
<head>
<title>READING XML FILES WITH AJAX</title>
<script type="text/javascript">
// initialize XMLHttpRequest object
var xmlobj=null;
var data=new Array();
// send http request
function sendRequest(doc){
    // check for existing requests
    if(xmlobj!=null&&xmlobj.readyState!=0&&xmlobj.readyState!=4){
        xmlobj.abort();
    }
    try{
        // instantiate object for Mozilla, Nestcape, etc.
        xmlobj=new XMLHttpRequest();
    }
    catch(e){
        try{
            // instantiate object for Internet Explorer
            xmlobj=new ActiveXObject('Microsoft.XMLHTTP');
        }
        catch(e){
            // Ajax is not supported by the browser
            xmlobj=null;
            return false;
        }
    }
    // assign state handler
    xmlobj.onreadystatechange=stateChecker;
    // open socket connection
  [b]  xmlobj.open('GET',doc,true); ===>access is denied. error [/b]
    // send GET request
    xmlobj.send(null);
}
// check request status
function stateChecker(){
    // if request is completed
    if(xmlobj.readyState==4){
        // if status == 200 display text file
        if(xmlobj.status==200){
            // create data container
            createDataContainer();
            // read XML data
            data=xmlobj.responseXML.getElementsByTagName
('song');
 // display XML data
            displayData();
        }
        else{
            alert('Failed to get response :'+ xmlobj.statusText);
        }
    }
}
// create data container
function createDataContainer(){
    var div=document.getElementById('container');
    if(div){return};
    var div=document.createElement('div');
    div.setAttribute('id','container');
    document.getElementsByTagName('body')[0].appendChild(div);
}
// display data at a given time interval
function displayData(){
    // reset data container
    document.getElementById('container').innerHTML='';
    var ul=document.createElement('ul');
    for(var i=0;i<data.length;i++){
        // create links
        var li=document.createElement('li');
        var a=document.createElement('a');
        // assign 'href' attribute
        a.setAttribute('href',data[i].getElementsByTagName('artist')
[0].firstChild.nodeValue);
        // add link labels
        a.appendChild(document.createTextNode(data
[i].getElementsByTagName('name')[0].firstChild.nodeValue));
        li.appendChild(a);
        ul.appendChild(li);
    }
    document.getElementById('container').appendChild(ul);
    // update headlines each 1 hour
    setTimeout("sendRequest('[URL unfurl="true"]http://www.radiojavan.com/rss.php?xml=true&count=10')",25*1000);[/URL]
}
// execute program when page is loaded
window.onload=function(){
    // check if browser is DOM compatible
    if(document.getElementById&&document.
getElementsByTagName&&document.createElement){
        // load XML file
        sendRequest('[URL unfurl="true"]http://www.radiojavan.com/rss.php?xml=true&count=10');[/URL]
    }
}
</script>
<style type="text/css">
#container {
    background: #eee;
    padding: 5px;
    border: 1px solid #000;
}
li {
    margin-top: 5px;
}
a:link,a:visited {
    font: bold 11px Tahoma, Arial, Helvetica, sans-serif;
    color: #00f;
}
a:hover {
    color: #f00;
}
</style>
</head>
<body>
</body>
</html>
 
Hi,

If the URL of the 'doc' you are trying to open is located on a different server, it will not work. AJAX has built in cross-domain scripting protection to prevent malicious use.

To get around this, you can use a scripting language like PHP to create a proxy page on your domain.

Code:
[blue]
<?php
  $feed = $_GET['feed'];
  $content = file_get_contents($feed);
  echo $content;
?>
[/blue]

This way, the domain is accessing the file and AJAX will allow this.

So, the parameter you would use for 'doc' would be:
Code:
[blue]
doc = '[URL unfurl="true"]http://yourdomain.com/proxy.php?feed=http://rss.cnn.com/rss/cnn_topstories.rss';[/URL]

[/blue]

Thats how I do it on my site and it works perfectly.

Hope that helps!
Ron





typedef map<GiantX,gold, less<std::shortestpathtogold> > AwesomeMap;
 
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