Xmpp 如何在Smack中正确侦听多用户聊天？

我不知道如何正确地监听多用户聊天，因为显然有两种不同的方式来监听它们

第一种方法是为每个房间添加MessageListener，如下所示：

MultiUserChat muc = new MultiUserChat(connection,"MyGroup");
muc.addMessageListener(new PacketListener() {
    @Override
    public void processPacket(Packet packet) throws NotConnectedException {
        // TODO Auto-generated method stub
    }
});

第二种方法是像收听普通的单人聊天信息一样收听：

PacketFilter filter = new MessageTypeFilter(Message.Type.groupchat);
connection.addPacketListener(new PacketListener() {
    @Override
    public void processPacket(Packet packet) {
        Message message = (Message) packet;
        if (message.getBody() != null) {
            final String fromName = StringUtils.parseBareAddress(message.getFrom());
            final String Rtext = message.getBody();

考虑到我也在监听文件传输，你能帮我找出哪种方式是正确的吗

MultiUserChat muc = new MultiUserChat(connection,"MyGroup"); muc.addMessageListener(new PacketListener() {
    @Override
    public void processPacket(Packet packet) throws NotConnectedException {
        // TODO Auto-generated method stub
    } });

---

是正确的方法。

因为原始代码都有bug，所以我使用了以下代码：

     PacketFilter messageFilter =

                new AndFilter(

                    new FromMatchesFilter(roomName+"@conference.reza-hp",true),

                    new MessageTypeFilter(Message.Type.groupchat));

            messageFilter = new AndFilter(messageFilter, new PacketFilter() {

                public boolean accept(Packet packet) {

                    Message msg = (Message) packet;

                    return msg.getBody() != null;



                }

            });

          addMessageListener(messageFilter);  
 }

    public static void addMessageListener(PacketFilter messageFilter) {

            connection.addPacketListener(new PacketListener() {

                @Override
                public void processPacket(Packet packet) throws NotConnectedException {
                     Message message = (Message)packet;  
                    String roomName = (message.getFrom().split("\\@conference.reza-
 hp"))[0];
                    String fromUser =    
 (StringUtils.parseResource(message.getFrom()).split("@\\@reza-hp"))[0];

                     System.out.println("=>"+"Room : "+roomName + "From   
 user:"+fromUser + ": "+message.getBody());  

                }
            }, messageFilter);



        }

---

在收到信息之前，我们需要加入房间吗？