ios - XMPPFramework - 如何将用户添加到 MUC 房间

Question

我正在创建群聊应用。我可以使用以下代码创建群组。

_xmppRoomStorage = [[XMPPRoomCoreDataStorage alloc]init];
   XMPPJID *roomJID = [XMPPJID jidWithString:@"room1@conference.abc.biz"];
   _xmppRoom =[[XMPPRoom alloc] initWithRoomStorage:_xmppRoomStorage jid:roomJID];
   [_xmppRoom              activate:_xmppStream];
   [_xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
   xmppRoom joinRoomUsingNickname:_userNameEdit.text history:nil];

但现在我需要向该组添加一些用户。谁能告诉我如何将多个用户添加或邀请到该组。

我还有一个问题。当第一组处于事件状态时无法创建第二个房间。当我尝试创建第二个房间时出现以下错误

“XMPPRoom[room2@conference.abc.biz] - 已经创建/加入/加入时无法创建/加入房间”

谢谢。 瓦兹

Answer 1

我通过以下方式解决了这个问题:

1. 先创建房间

   -(void) CreateRoom
   {
       XMPPJID *roomRealJid = [XMPPJID jidWithString:jidRoom];// Room name ex. abc@conference.xyz.biz
       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. 发送邀请

   // 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"];
           }
       }
   }
   

希望这对您有所帮助。