Ext 获取返回数据问题

Question

初学Ext，想做一个简单的登陆如下
loginForm.doAction('submit',{
url:'user.action',
method:'POST',
waitMsg:'正在登陆....',
success:function(form,action){
  alert(action.result.msg);
var isSuc = action.result.success;  
          if(isSuc) { Ext.Msg.alert('消息', '登陆成功..'); };
        },
        failure:function(form, action){  
            alert('登陆失败');  
        }
})后台java登陆成功返回了数据的，但不知道前台怎么接收。上面的不对，怎么接收？
部分后台代码
public String userCheck()throws Exception{
User u = new User();
u.setUsername(info.getUsername());
u.setPassword(info.getPassword());
System.out.println("NAME :"+u.getUsername()+" password :"+u.getPassword());
if(userService.checkUserPassword(u))
return SUCCESS;
return "fail";
}
struts2 代码<action name="user" class="com.boco.web.action.UserAction" method="userCheck">

Answer 1

后台不是这样返回的
要用response write出来不能直接return success

Answer 2

后台的代码不能return 因为你是异步请求 响应必然要到前台去 不论结果如何都要向前台写
struts.xml中不需要配置resault了 只需要配置其对应的action就可以了
if(userService.checkUserPassword(u)){
response.getWriter().write('成功');//response哪来的 楼主你懂得
}
response.getWriter().write('失败')

Answer 3

public void GetPersonnelList(HttpServletResponse response){
        System.out.println("获取员工列表!");
        PersonnelDao dao = new PersonnelDao();
        List list = dao.getPersonnelList();
        try{
            /*
             * json 需要导入包: 
             * json-lib-2[1].2.2-jdk15.jar 
             * commons-collections.jar 
             * commons-digester.jar commons-lang.jar
             * ezmorph-1[1].0.4.jar
             * */
            JSONArray jsonArray = JSONArray.fromObject(list);            
            sendJson(response, jsonArray.toString());            
        }catch(Exception e){
            e.getMessage();
        }
    }
public void sendJson(HttpServletResponse response,String jsonStr){
        PrintWriter out = null;
        try{
            out = response.getWriter();
            out = response.getWriter();
            System.out.println(jsonStr);
            out.print(jsonStr);
            out.flush();
            out.close();
        }catch(Exception e ){
            e.printStackTrace();
        }finally{
            out.flush();
            out.close();    
        }
            }
Ext.Ajax.request({ //ajax异步请求后台 
            url: '/Extjs_Center/personnelAction.do?method=personnelList', 
            method:'post', 
            success:function(response,opt){             
                var data = response.responseText;        //返回后台json数据                     
                var results= Ext.decode(data);            
                store.proxy = new Ext.ux.data.PagingMemoryProxy(results); //PagingMemoryProxy() 一次性读取数据 
                store.load({params:{start:0,limit:13}});        //导入后数据显示
            }, 
            failure: function(){
                Ext.Msg.alert("failure");
            } 
        });

Answer 4

public void GetPersonnelList(HttpServletResponse response) 跟了参数，怎么找不到这个方法呀？要怎么设置呢？

Answer 5

我倒啊~~~我说楼主你懂得 看来你不懂 跟参数干嘛
在方法里边直接用这个
 
if(userService.checkUserPassword(u)){
response.getWriter().write('成功');//HttpServletResponse response 就这么直接用
}
response.getWriter().write('失败')