json数据娶不到值。。大神指教

Question

前台页面

            <li>
            <div class="fc">
              <input name="id" type="hidden" id="id"/>                       
            </div>
          </li>

          <li>
            <label class="label">用户名：<span>*</span></label>
             <div class="fc">
              <input name="name" type="text" maxlength="40" id="name" value="${u.name}" />  <span class="tips">最多40个字符</span>
             </div>
          </li>

          <li>
           <label class="label">邮箱：<span>*</span></label>
            <div class="fc">
             <input name="email" type="text" maxlength="16" id="email" value="${u.email}" /> <span class="tips">最多16个字符</span>
            </div>

                        <script type="text/javascript">

function Preservation(){
alert("点击保存");
var name = $("#name").val();
alert("----用户信息---"+name);
var email = $("#email").val();
alert("---用户邮箱----"+email);
var id = ${u.id}
alert("---用户ID----"+id);
$.ajax({
url:"<%=request.getContextPath()%>/login_update.action",
type:"post",
data:{
"id":id,
"name":name,
"email":email
},
dataType:"json",
success:function(obj){
if(obj){
alert("修改成功！");
location.href = "<%=request.getContextPath()%>/login_toMain.action;
}
else{
alert("修改失败！");
}

}
})

}

后台接收不到值，更新不了数据，/**
* 修改action
* @return
* @throws IOException
* @throws SQLException
*/
public String update() throws SQLException {
//通过ID找对象
UserInfo userInfo = UserInfoDataProvider.getInstance().getUserInfoById(id);
System.out.println("============"+userInfo.getName());
System.out.println("----------------"+userInfo.getId());
//更新
UserInfoDataProvider.getInstance().updateUserInfo(userInfo);

    HttpServletResponse response = ServletActionContext.getResponse();

    JSONObject json = JSONObject.fromObject(userInfo);

    try {
        response.getWriter().write(json.toString());
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    return null;

Answer 1

你用request取id看有不呢