无法使用 smack-android:4.1.4 接收群聊消息 [英] Can't able to receive group chat messages using smack-android:4.1.4

问题描述

我正在使用 smack 库开发聊天应用程序.我在群聊中遇到问题.在我的应用程序中，我正在创建一个组，并且该组成员是自动加入的.即使他们没有发起聊天，我也想在组中发送消息时通知所有用户.我的代码如下所示将侦听器置于 init 方法中但无法接收消息.

I am developing chat app using smack library. I have an issue in group chat. In my app, i am creating a group and in that members are auto-joined.i want to notify all user when I send a message in the group even if they had not initiated a chat.My code is as follow in that I have place listener in init method but unable to receive a message.

        multiUserChatManager = MultiUserChatManager.getInstanceFor(mConnection);
        mMultiUserChat = multiUserChatManager.getMultiUserChat(to);
        mConnection.addAsyncStanzaListener(this, null);
        DiscussionHistory history = new DiscussionHistory();
        history.setMaxStanzas(0);
        mMultiUserChat.addMessageListener(this);
        mConnection.addSyncStanzaListener(this, null);
        try {
            mMultiUserChat.join(from, "", history, SmackConfiguration.getDefaultPacketReplyTimeout());
        } catch (SmackException.NoResponseException e) {
            e.printStackTrace();
        } catch (XMPPException.XMPPErrorException e) {
            e.printStackTrace();
        } catch (SmackException.NotConnectedException e) {
            e.printStackTrace();
        }

这里是群的消息监听器

 @Override
public void processMessage(Message message) { 
    Logg.e(TAG,"Message received group.."); 
}

我不知道为什么当有人在组中发送消息时此方法不调用，即使我加入了组，如果我创建了1个组并加入了2个用户，当1个用户在组中发送消息时，user2无法接收消息，但是当用户 2 在该组内发送消息时，他们都可以接收消息.

I don't know why this method does not call when someone send message in group, even I joined group, If I create 1 group and joined 2 users, when 1 user sends message in group then user2 can't able to receive message, but when user 2 send message inside this group then they both are able to receive messages.

请帮助我，我无法找到解决方案.请不要给出已经过时的建议.

Please help me, I can't able to find the solution. Please don't give suggestion which is already deprecated.

提前致谢.！！

推荐答案

在完整的代码审查后，我正在完整地编辑答案.-再次-

我建议重构您的代码，以在 1 个以上的大类中保持角色分离.

I suggest to refactor your code to keep separation of roles in more than 1 huge class.

基本上，由于您的代码中有许多addasync - addsync"，您在错误的侦听器中看到消息，并且您能够接收消息，就像您的怪物级全入的副作用一样！

Basically you are especting messages in wrong listener due to many "addasync - addsync" in your code and you are able to receive messages just as side effect of your monster-class-all-in!

我看到您需要对代码应用许多优化.解释太长，不可能，但是，举个例子:

I see many optimization you need to apply to your code. It's too long to explain and out of the question, however, just as example:

1. sendGroupMessage You can check by MultiUserChatManager if you
already joined the chat and then send the message. You must fire a
"join" just once, not everytime you want to send a message.

2.mMultiUserChat.addMessageListener(this);必须添加一次侦听器，否则您将创建大量线程.可能它有效，因为你有一个单身人士.当您有一个听众时，如果您不将其删除，则无需再将其添加到该聊天中.

2. mMultiUserChat.addMessageListener(this); A listener must be added ONCE or you'll create tons of threads. Probably it works because you have a singleton. While you have a listener, you don't need to add it anymore to that chat if you don't remove it.

1. mConnection.addSyncStanzaListener(this, null);小心:您正在将您的侦听器(只有一个?您实现了大量具有相同类的侦听器)添加到您的连接中.在您的代码之前或之后，您的代码会吃掉一个重要的节(prolly 自定义 IQ)，并且您将很难发现副作用.
2. mConnection.addAsyncStanzaListener(this, null); 与 3 相同
3. 在之前或之后检查 ProviderManager.addExtensionProvider()你需要一些.

1. mConnection.addSyncStanzaListener(this, null); Be carefull: you are adding your listener (wich one? You implements tons of listeners with same class) to your connection. Before or later your code will eat an important stanza (prolly a custom IQ) and you'll have an hard to discovery side effects.
2. mConnection.addAsyncStanzaListener(this, null); same of 3
3. Check for ProviderManager.addExtensionProvider(), before or later you'll need some.

希望有所帮助.

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