关于javascript：无法将JSONObject从JAVA Class传递给jsp

Unable to pass JSONObject from JAVA Class to jsp

我已经看过这个例子，但是我仍然无法在JSP中检索JSON对象。以下是myCalendarController.java类中的代码：

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public class MyCalendarController implements Controller{

public ModelAndView handleRequest(HttpServletRequest request,
HttpServletResponse response) throws Exception {

if("Add".equals(request.getParameter("action"))){
...
JSONObject jObj = new JSONObject();
jObj.put("test","Success");
response.getWriter().write(jObj.toString());
...
}
return new ModelAndView("mycalendar","model", myModel);
}

下面是我试图在JSP中检索它的方法，但是警报总是说"未定义"。

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var queryString ="?action=Add";
queryString += "&t=" + title;
queryString += "&sDT=" + stDate +"T" + stHour +":" + stMin +":00";
queryString += "&eDT=" + eDate +"T" + eHour +":" + eMin +":00";
$.ajax({
type:"GET",
url:"mycalendar.htm" + queryString,
success: function(response){
alert(response.test);
}
});

注意：当从JSP对类进行Ajax调用时，我正在尝试创建JSON对象。我不熟悉Ajax和Javascript，所以一定是做错了什么…请帮助！！！！

在上述代码中，response.responseText属性为"未定义"。但我尝试了另一种方法：

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var ajaxRequest;

try{
ajaxRequest = new XMLHttpRequest();
}catch (e){
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
alert("Your browser broke!");
return false;
}
}
}

ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
alert(ajaxRequest.responseText);
alert("test:" + ajaxRequest.test);
}
}

var queryString ="?action=Add";
queryString += "&t=" + title;
queryString += "&sDT=" + stDate +"T" + stHour +":" + stMin +":00";
queryString += "&eDT=" + eDate +"T" + eHour +":" + eMin +":00";

ajaxRequest.open("GET","mycalendar.htm" + queryString, true);
ajaxRequest.send(null);

这样我就得到了AjaxRequest.ResponseText，但是警报("test："+AjaxRequest.test)；仍然显示未定义

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var a = '<%=DropDown.returnList()%>';
var countryObj = JSON.parse(a);
var s = $('#selectCountry');
for(var val in countryObj)
{
$('<option />', {value: val, text: countryObj[val]}).appendTo(s);
}