jquery$.ajax()及其成功回调-返回到哪里?

我编写了一个请求方法,该方法执行ajax调用,检测我是否提供了回调或返回响应XML的datanode。

function request(request, dontRefresh)
{     
   var requestXML = composeRequestXML(request); 
   $.ajax({
      url: 'someProcessor.php',
      type: 'POST',
      cache: false,
      async: dontRefresh,
      timeout: 5000,
      data: "query="+requestXML,
      success: function(response)
      {
         //parses xml into a js object
         var responseObj = parseResponseXML(response);

         if(request.callback){
            request.callback(responseObj);
         } else {
            // responseObj.response[0].data[0] is the data 
            // node of the response Obj.
            // this is what i see being the problem - 
            // I DON'T KNOW WHERE THIS IS RETURNED TO!!
            return responseObj.response[0].data[0];
         }
      }
   });
}

此请求将使用回调

var requestObj = new Object();
    requestObj.callback = function(responseObj){someCallbackFunction(responseObj);};
    requestObj[0] = new Object();
    requestObj[0].module = "someModule";
    requestObj[0].action = "someModuleMethod";
request(requestObj);
//results are returned to the function someCallbackFunction()

这是我想完成的一个例子

var requestObj = new Object();
    requestObj[0] = new Object();
    requestObj[0].module = "userManager";
    requestObj[0].action = "GET_USERID";

var userId = request(requestObj, false); //make the request asynchronous

我试过把 $.ajax()

function request(request, dontRefresh){
   return $.ajax({/*...options...*/);
}

但这绕过了我开发的xml解析器,返回了XHR对象。我想使用这种技术来注册变量。所以本质上。。。

我将对回调使用这个方法,或者用它设置一个变量。