一个线程中的两个message如何处理？

Java code

 //获取好友分组的线程       new Thread(new Runnable() {            @Override            public void run() {                // TODO Auto-generated method stub                Map<String, Object> params = new HashMap<String, Object>();                map = new HashMap<String, Object>();                friendGroup = new ArrayList<FriendGroup>();                                     friendList = new ArrayList<FriendGroup>();                params.put("ID", LoginActivity.userID);                //获取用户组的输入流                InputStream group_is = WebServer.hcPostData("GetFriendGroup",                        "FriendApp.ashx", params);                                     InputStream friends_is = WebServer.hcPostData("GetFriendList",                        "FriendApp.ashx", params);                parser = new XMLParser();                try {                    friendGroup = parser.FriendGroupParser(group_is);                                              friendList = parser.FriendGroupParser(group_is);                    Message message = new Message();                    message.obj = friendGroup;                    mHandler.sendMessage(message);                } catch (Exception e) {                    // TODO Auto-generated catch block                    e.printStackTrace();                }            }        }).start();               private Handler mHandler = new Handler() {        @SuppressWarnings("unchecked")        @Override        public void handleMessage(Message msg) {                        // TODO Auto-generated method stub            friendGroup = (ArrayList<FriendGroup>) msg.obj;             GetAllFriend();            super.handleMessage(msg);        }    };

这是我想到的，但是一个handler只能处理一个message啊！但是我有两个对象obj，一个FriendGroup，一个FriendList，只能处理一个怎么办呢？

[解决办法]
加个
message.what = 1;
及
message.what = 2;

然后在
handleMessage

switch(msg.what)
{
case 1:
.....
break;
case 2:
.....
break;
}



[解决办法]
定义两个消息，即what

Java code

static final int SEND_GROUP = 1;static final int SEND_LIST = 2;