帮我看看下面的ajax代码有什么问题？

Question

$(document).ready(function(){
$.ajax({
type:"post",
url:"friendAction!searchUI.action",
data:"<%=request.getParameter("userId") %>",
async:false,
success:function(msg){
alert(msg);
}
});
})
//辅助获取跳转到的页面信息
public ActionForward searchUI(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response) {

    //得到好友
    Integer userId = Integer.valueOf(request.getParameter("userId"));

    userList = userService.getResult("from Users where id=?", new Object[]{userId});

    return mapping.findForward(null);
}
已经为userList设了get和set方法。

${userList.name } 打不出值。

Answer 1

ajax请求非异步返回通常返回json类型，然后在success分支里面获取返回的数据。
你是不是想把userList值返回给ajax的回调函数呢？如果是的话，可以这样修改js和action方法，示例代码：

 $.ajax({
type:"post",
dataType:"json",
url:"friendAction!searchUI.action",
data:"<%=request.getParameter("userId") %>",
async:false,
success:function(msg){
alert(msg);
}
});
})

Action中将数据转成JSON返回。

 response.setContentType("text/plain");// 设置输出为文字流
response.setCharacterEncoding("UTF-8");
PrintWriter out = null;
try {
    out = response.getWriter();
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
out.write(JSON.toJSONString(obj));
out.flush();
out.close();