通过soap通讯后,如何获得返回的xml属性值？？

<?xml version="1.0" encoding="GB2312" standalone="yes"?>
<Msg Version="4" MsgID="1104911830" Type="TVMonUp" DateTime="2005-01-05 15:57:10
" SrcCode="110000N01" DstCode="900000X10" ReplyID="37041743">
<Return Type="StreamSimpleQuery" Value="0" Desc="成功" />
<stream>
  <mediastream index="0" url="HTTP://192.168.0.2/235364F_0_41DB9DD5.asx" />
</stream>
</Msg>

如何获得url的值？？
if(xmlhttp.status==200)
{
  var xmlDOC = new ActiveXObject("MSXML.DOMDocument");
  xmlDOC.load(xmlhttp.responseXML);
  Number_Data=xmlDOC.documentElement.selectNodes("//url")(0).text;
alert(RecordNumber);
}
请教高手？？

alert(xmlhttp.responseXML.xml) 的结果是什么?