XMPPFramework - 如何将用户添加到 MUC Room答案

【问题标题】：XMPPFramework - How to Add Users to MUC RoomXMPPFramework - 如何将用户添加到 MUC Room
【发布时间】：2013-09-04 13:14:37
【问题描述】：

我正在创建群聊应用。我可以使用以下代码创建组。

_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];

但现在我需要将一些用户添加到该组。谁能告诉我如何添加或邀请多个用户到这个组。

我还有一个问题。第一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，它给出了以下错误

“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

【问题讨论】：

我已经解决了这个问题......
你能分享一下我实际上面临同样问题的解决方案

标签： ios objective-c xmpp xmppframework

【解决方案1】：

我通过以下方式解决了这个问题：

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]];
}

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"];
        }
    }
}

希望这会有所帮助。

【讨论】：