求XPath表达式（怎样根据一个节点的属性来找到该节点）

Question

http://topic.csdn.net/t/20060706/14/4864092.html

Answer 1

function changeNodeValue()
        {
           var dom=new ActiveXObject("MSXML2.DOMDocument.5.0"); 
           dom.loadXML("<ControlData><control id=\"control1\" value=\"AA\"/><control id=\"control2\" value=\"BB\"/></ControlData>");
           var node=dom.selectSingleNode("/ControlData/control[@id='control2']");
           if (node)
           {
                node.setAttribute("value","bb")
           }
           window.alert(dom.xml);
        }

Answer 2

//control[@id="control2"]

== 思想重于技巧 ==

  

Answer 3

nodeTop.SelectSingleNode("//control[@id="control2"]").SetAttribute("value")="BB";

== 思想重于技巧 ==

  

Answer 4

var find=function() 
        { 
           var Dom=new ActiveXObject("MSXML2.DOMDocument.5.0");  
           Dom.loadXML(" <ControlData> <control id=\"control1\" value=\"AA\"/> 
           <control  id=\"control1\" value=\"BB\"/> </ControlData>"); 
           var node=dom.selectSingleNode("/ControlData/control[@id='control1']"); 
           node.setAttribute("value","bb")           
           alert(Dom.xml); 
        };