XMPPFramework - 如何将用户添加到MUC聊天室

Question

我正在创建一个群聊应用程序。使用以下代码可以创建群组。

_xmppRoomStorage = [[XMPPRoomCoreDataStorage alloc]init];
   XMPPJID *roomJID = [XMPPJID jidWithString:@"[e-mail]"];
   _xmppRoom =[[XMPPRoom alloc] initWithRoomStorage:_xmppRoomStorage jid:roomJID];
   [_xmppRoom              activate:_xmppStream];
   [_xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
   xmppRoom joinRoomUsingNickname:_userNameEdit.text history:nil];

现在我需要在这个群组中添加一些用户。请问有人能告诉我如何添加或邀请多个用户到这个群组吗？

我还有一个问题，当第一个群组处于活动状态时，无法创建第二个房间。当我尝试创建第二个房间时，会出现以下错误：

"XMPPRoom[[e-mail]] - 已经在创建/加入/已加入房间时无法再次创建/加入房间"

谢谢。 Vaz

- Vaz

---

我已经解决了这个问题...... - Vaz

你能分享一下解决方案吗？我遇到了同样的问题。 - Azerue

Answer 1

我通过以下方式解决了这个问题：

1. Create the room first

   -(void) CreateRoom
   {
       XMPPJID *roomRealJid = [XMPPJID jidWithString:jidRoom];// Room name ex. [e-mail]
       XMPPRoom *newXmppRoom = [[XMPPRoom alloc] initWithRoomStorage:[[self appDelegate] xmppRoomStorage] jid:roomRealJid    dispatchQueue:dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0)];
       [newXmppRoom activate: [[self appDelegate] xmppStream]];
       [newXmppRoom fetchConfigurationForm];
       [newXmppRoom addDelegate:[self appDelegate] delegateQueue:dispatch_get_main_queue()];
       [newXmppRoom joinRoomUsingNickname:nickName history:nil password:[[NSUserDefaults     standardUserDefaults] stringForKey:kXMPPmyPassword]];
   }
   

2. Send invitations

   // Once the room created, we get some responses from server. 
   // One of them is "didFetchModeratorsList".
   
   - (void)xmppRoom:(XMPPRoom *)sender didFetchModeratorsList:(NSArray *)items
   {
       DDLogInfo(@"%@: %@ --- %@", THIS_FILE, THIS_METHOD, sender.roomJID.bare);
   
       if (check the flag for room create and invite) // This has to be done only when we intended
       {
           NSArray* users = list of users we need to invite.
   
           if (users.count > 0)
           {
               for (int i=0; i< users.count; i++)
               {
                   NSString *jid = [NSString stringWithFormat:@"%@@xyz.biz", [users objectAtIndex:i]];
                   XMPPJID *xmppJID=[XMPPJID jidWithString:jid];
                   [sender inviteUser:xmppJID withMessage:@"Join Group."];
               }
               [sender sendMessageWithBody:@"Hi All"];
           }
       }
   }
   

希望这能帮到你。

- Vaz