This is just an example. I only did it for firefox and it depends on only having three entrees in xml. I hope you know how to read an xml and get the right url:
The XML:
Code:
<?xml version="1.0" encoding="ISO-8859-1"?>
<links>
<link category="COOKING">
<url>goole</url>
<img>2005</img>
</link>
<link category="CHILDREN">
<url>google</url>
<img>2005</img>
<price>29.99</price>
</link>
<link category="WEB">
<url>http://facebook.com/</url>
<img>2003</img>
</link>
</links>
The Javascript
:
Code:
<!DOCTYPE html>
<html>
<body>
<script>
function loadXMLDoc(dname)
{
if (window.XMLHttpRequest)
{
xhttp=new XMLHttpRequest();
}
else
{
xhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xhttp.open("GET",dname,false);
xhttp.send("");
return xhttp.responseXML;
}
xml=loadXMLDoc("Links.xml");
path="/links/link/url"
// code for IE
if (window.ActiveXObject)
{
var nodes=xml.selectNodes(path);
for (i=0;i<nodes.length;i++)
{
document.write(nodes[i].childNodes[0].nodeValue);
document.write("<br>");
}
}
// code for Mozilla, Firefox, Opera, etc.
else if (document.implementation && document.implementation.createDocument)
{
var nodes=xml.evaluate(path, xml, null, XPathResult.ANY_TYPE, null);
var result=nodes.iterateNext();
while (result)
{
jimmy = result.childNodes[0].nodeValue;
document.write(result.childNodes[0].nodeValue);
document.write("<br>");
result=nodes.iterateNext();
}
}
/*bill = String(jimmy); <= YOU MIGHT HAVE TO DO THIS, DON'T KNOW*/
window.open(jimmy);
</script>
</body>
</html>