XMPPFramework - 如何创建MUC房间并邀请用户？

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房间，但它不起作用。

我正在使用以下代码：

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[email protected] /cdn-cgi/l/email-protection/room" nickName:@"room"];
[room createOrJoinRoom];
[room sendInstantRoomConfig];
[room setInvitedUser:@"[email protected] /cdn-cgi/l/email-protection"];
[room activate:[self xmppStream]];    
[room inviteUser:jid1 withMessage:@"hello please join."];
[room sendMessage:@"HELLO"];

用户[电子邮件受保护] /cdn-cgi/l/email-protection应该会收到邀请消息，但什么也没发生。

任何帮助将不胜感激。 :)

在探索了各种解决方案之后，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. [email protected] /cdn-cgi/l/email-protection
    */
   
   XMPPJID *roomJID = [XMPPJID jidWithString:@"[email protected] /cdn-cgi/l/email-protection"];
   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. 检查该委托中的房间是否已成功创建：

   - (void)xmppRoomDidCreate:(XMPPRoom *)sender
   

3. 检查您是否已加入该代表的房间：

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender
   

4. 创建房间后，获取房间配置表：

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

5. 配置您的房间

   /**
    * 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：多用户聊天 http://xmpp.org/extensions/xep-0045.html#createroom-reserved, 实施群聊 https://stackoverflow.com/questions/19268629/xmpp-ios-chat-client-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/群组聊天室，并邀请了一个用户。 :)