ajax异步请求返回对象

使用ajax异步请求返回一个对象。

java code:

 @RequestMapping({"getAstSingleWheelImg_bbs"+Constant.JSON})
  @ResponseBody
  public Result getImgUrl(HttpServletRequest request, Model model, WheelChart chart)throws ParseException{
 String userName = request.getParameter("userName");
    System.out.println("userName:"+userName);
    String astroDate = request.getParameter("astroDate");
    String astroHour = request.getParameter("astroHour");
    String astroMin = request.getParameter("astroMin");
    //略过部分代码

    result.setSuccess(true);
    result.setDesc(show_imgUrl);
    System.out.println(show_imgUrl);
    return result;
}

js code:

var a=$("#wheelImg");
            $.ajax({
                async:true,
                type:"post",
                contentType:"application/x-www-form-urlencoded",
                cache:false,
                url:request_url+"/getAstSingleWheelImg_bbs.jo",
                data:a.serializeArray(),
                dataType:"json",
                timeout:6000,
                beforeSend:function () {
                    alert("正在处理请求，请稍后。。。。。");
                },
                success:function(result){
                    $("img_wheel").attr("src",result.desc);
                    alert("success");
                },
                error:function(XMLHttpRequest, textStatus, errorThrown){
                    alert(XMLHttpRequest.status);
                    alert(XMLHttpRequest.readyState);
                    alert(textStatus);
                }
            });

　　不解释太多了，网上的资料一堆堆的解释ajax。我就backup code..ok..