ajax接收不到action的值！

ajax接收不到action的值！求助！
我们刚学习了自定义MVC框架。ajax遇到问题
页面接受不到action的值。

if(XMLHTTPResult.readyState==4 && XMLHTTPResult.status==200){
这一行代码也不走。

求高手援助。。

js代码

JScript code

function getStreet(){    var district = document.getElementById("district_id");    var url = "getStreet.action?districtId="+district.value;    createXmlHttpResult();    XMLHTTPResult.open("GET", url,true);    XMLHTTPResult.onreadystatechange=backRoll;    XMLHTTPResult.send(null);}function backRoll(){    if(XMLHTTPResult.readyState==4){        if(XMLHTTPResult.status==200){            alert(XMLHTTPResult.responseText);                           }    }        }

XML code

 <action name="getStreet" class="com.accp.houserent.action.GetStreetAction">     </action>

Java code

public String execute(HttpServletRequest request,            HttpServletResponse response) throws Exception {        // TODO Auto-generated method stub        String districtId = request.getParameter("districtId");        CommonBiz commonBiz = new CommonBizImpl();        District district =commonBiz.getEntity(District.class,Integer.parseInt(districtId));        response.setContentType("html/text;charset=gbk;");        PrintWriter out = response.getWriter();        String str  = null;        for (Iterator<Street> ite = district.getStreets().iterator(); ite.hasNext();) {            Street temp_street  = ite.next();            if (ite.hasNext()) {                str+=temp_street.getStreetName()+"-";            }else{                str+=temp_street.getStreetName();            }        }        out.print(str);        return null;    }

JScript code

function backRoll(){    if(XMLHTTPResult.readyState==4){        if(XMLHTTPResult.status==200){            alert(XMLHTTPResult.responseText);                           }        else alert('发生错误：'+XMLHTTPResult.status+'\n'+XMLHTTPResult.responseText)//加这句看看什么问题    }        }