同一段代码，不同方式访问结果不一样，请赐教！

Hotlion 2009-07-20 03:40:21

以下是一个HTML文件，如果直接双击打开，可正常显示，但如果是通过网站地址打开的话，就会提示出错“Line:27
Char:3
Code:0
Error:拒绝访问。”
查了一下，应该是变量xmlsource在xmlDoc.load(xmlsource)这一句中出了问题，请问如何纠正这个问题，谢谢！

    <div id="container" >News Loading...</div>

<script language="JavaScript1.2">

 //Container for ticker. Modify its STYLE attribute to customize style:

 var tickercontainer=''

 var xmlsource="http://api.jiwai.de/statuses/user_timeline/408742.rss";

 var root;

 var title;

 var link; 

 var items;

 var item;

 var images;

 var image;

 var description;

 if (window.ActiveXObject)

 {

  //document.write("Microsoft.XMLDOM");

  var xmlDoc = new ActiveXObject("Microsoft.XMLDOM");

 }

 else if (document.implementation && document.implementation.createDocument)

 {

  //document.write("document.implementation.createDocument");

  var xmlDoc= document.implementation.createDocument("","doc",null);

 }

 if (typeof xmlDoc!="undefined")

 {

  //document.write(tickercontainer)

  xmlDoc.load(xmlsource)

 }  

 function fetchxml()

 {

  if (xmlDoc.readyState==4)

   output()

  else

   setTimeout("fetchxml()",10)

 }

 function output()

 {  

  var temp="";

  root = xmlDoc.getElementsByTagName("channel")[0];

  title =root.getElementsByTagName("title")[0];

  //temp = title.firstChild.nodeValue +"<br>";

  items=root.getElementsByTagName("item");

  for(i=0;i<=items.length-1;i++)

  {

   item=items[i];

   title=item.getElementsByTagName("title")[0]; 

   link=item.getElementsByTagName("link")[0];

   description=item.getElementsByTagName("description")[0];

   temp = temp + "<font size=-1><a href=" + link.firstChild.nodeValue+ " target='_bank'>" + title.firstChild.nodeValue +"</a></font><br><br>";

   //temp = temp +"<font size=-1>" + description.firstChild.nodeValue + "</font><br><br>";

   document.getElementById("container").innerHTML = temp;

  }

 }

 if (window.ActiveXObject)

  fetchxml()

 else if (typeof xmlDoc!="undefined")

  xmlDoc.onload=output

 </script>

</body>

</html>