xml文件是
<?xml version="1.0" encoding="GB2312"?>
<item>
<class id="netease">
<subclass url="http://www.163.com"></subclass>
</class>
<class id="sohu">
<subclass url="http://www.sohu.com"></subclass>
</class>
</item>向读取xml 比如readxml('sohu');函数,使用这样函数直接打开http://www.sohu.com<a href="#" onclick="readxml('sohu')">点击</a>点击链接,打开http://www.sohu.com这样的js应该怎样去考虑那?
<?xml version="1.0" encoding="GB2312"?>
<item>
<class id="netease">
<subclass url="http://www.163.com"></subclass>
</class>
<class id="sohu">
<subclass url="http://www.sohu.com"></subclass>
</class>
</item>向读取xml 比如readxml('sohu');函数,使用这样函数直接打开http://www.sohu.com<a href="#" onclick="readxml('sohu')">点击</a>点击链接,打开http://www.sohu.com这样的js应该怎样去考虑那?
------------------------
<?xml version="1.0"?>
<item>
<class id="netease">
<subclass url="http://www.163.com"></subclass>
</class>
<class id="sohu">
<subclass url="http://www.sohu.com"></subclass>
</class>
</item>
-----------<script>
var loaded = false;
var oDoc = new ActiveXObject("Msxml2.DOMDocument");
oDoc.async = true;
oDoc.onreadystatechange = doload;
oDoc.load("xml.xml");function doload(){
if(oDoc.readyState==4){loaded = true;}
}function readxml(n){
if(loaded){
var oNode = oDoc.selectSingleNode("//class[@id='" + n + "']/subclass");
var s = oNode.getAttribute("url");
location.href = s;
}
}</script>
<a href="#" onclick="readxml('sohu')">点击</a>
<script>
var loaded = false;
var oDoc = new ActiveXObject("Msxml2.DOMDocument");
oDoc.async = true;
oDoc.onreadystatechange = doload;
oDoc.load("xml.xml");
function doload(){
if(oDoc.readyState==4){
if(oDoc.parseError.errorCode!=0) {
alert("Error:" + oDoc.parseError.reason);
return false;
}
else loaded = true;
}
}function readxml(n){
if(loaded){
var oNode = oDoc.selectSingleNode("//class[@id='" + n + "']/subclass");
var s = oNode.getAttribute("url");
location.href = s;
}
}</script>
<a href="#" onclick="readxml('sohu')">点击</a>
在这处理成打开一个新链接,应该怎么处理啊
window.open(s);