XMPPFramework - 如何创建 MUC 房间并邀请用户？

Question

我正在使用 Robbiehanson 的 iOS XMPPFramework。我正在尝试创建一个 MUC 房间并邀请用户加入群聊室，但它不起作用。

我正在使用以下代码：

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"user101@conference.jabber.org/room" nickName:@"room"];
[room createOrJoinRoom];
[room sendInstantRoomConfig];
[room setInvitedUser:@"ABC@jabber.org"];
[room activate:[self xmppStream]];    
[room inviteUser:jid1 withMessage:@"hello please join."];
[room sendMessage:@"HELLO"];

用户 ABC@jabber.org 应该会收到邀请消息，但没有任何反应。

任何帮助将不胜感激。 :)

Answer 1

在探索了各种解决方案后，我决定在这里编译并分享我的实现：

1. 创建 XMPP 房间：

   XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];
   
   /** 
    * Remember to add 'conference' in your JID like this:
    * e.g. uniqueRoomJID@conference.yourserverdomain
    */
   
   XMPPJID *roomJID = [XMPPJID jidWithString:@"chat@conference.shakespeare"];
   XMPPRoom *xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomStorage
                                                          jid:roomJID
                                                dispatchQueue:dispatch_get_main_queue()];
   
   [xmppRoom activate:[self appDelegate].xmppStream];
   [xmppRoom addDelegate:self 
           delegateQueue:dispatch_get_main_queue()];
   
   [xmppRoom joinRoomUsingNickname:[self appDelegate].xmppStream.myJID.user 
                           history:nil 
                          password:nil];
   

2. 检查此委托中是否成功创建房间：

   - (void)xmppRoomDidCreate:(XMPPRoom *)sender
   

3. 检查您是否已加入此代表的房间：

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender
   

4. 房间创建完成后，获取房间配置表单：

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender {
       [sender fetchConfigurationForm];
   }
   

5. 配置你的房间

   /**
    * Necessary to prevent this message: 
    * "This room is locked from entry until configuration is confirmed."
    */
   
   - (void)xmppRoom:(XMPPRoom *)sender didFetchConfigurationForm:(NSXMLElement *)configForm 
   {
       NSXMLElement *newConfig = [configForm copy];
       NSArray *fields = [newConfig elementsForName:@"field"];
   
       for (NSXMLElement *field in fields) 
       {
           NSString *var = [field attributeStringValueForName:@"var"];
           // Make Room Persistent
           if ([var isEqualToString:@"muc#roomconfig_persistentroom"]) {
               [field removeChildAtIndex:0];
               [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]];
           }
       }
   
       [sender configureRoomUsingOptions:newConfig];
   }
   

   参考：XEP-0045: Multi-User Chat、Implement Group Chat

6. 邀请用户

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender 
   {
       /** 
        * You can read from an array containing participants in a for-loop 
        * and send multiple invites in the same way here
        */
   
       [sender inviteUser:[XMPPJID jidWithString:@"keithoys"] withMessage:@"Greetings!"];
   }
   

在那里，您创建了一个 XMPP 多用户/群聊室，并邀请了一位用户。 :)

Answer 2

我感觉alloc-init之后要做的第一件事就是把它附加到你的xmppStream上，这样它就可以使用xmppStream来发送/接收消息了。

更准确地说：

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"user101@conference.jabber.org/room" nickName:@"room"];
[room activate:[self xmppStream]];

//other things (create/config/...)

Answer 3

查看最新的 XMPPMUCLight 和 XMPPRoomLight，它类似于 Whatsapp 和其他当今趋势的社交应用程序房间，它们在离线或房间内无人时不会被破坏或成员被踢。

参考这个documentation & mod from MongooseIM