无法让成员进入房间 - XMPP答案

【问题标题】：Not able to get members in a room - XMPP无法让成员进入房间 - XMPP
【发布时间】：2018-09-20 07:36:48
【问题描述】：

我正在获取所有组的名称...

  func xmppMUC(_ sender: XMPPMUC, didDiscoverRooms rooms: [Any], forServiceNamed serviceName: String) {
    if let elements = rooms as? [DDXMLElement] {
      for element in elements {

        print("Name: \(String(describing: element.attributeStringValue(forName: "name")))")
        print("JID:  \(String(describing: element.attributeStringValue(forName: "jid")))")

      }
    }
    print("rooms: \(rooms)")

}

这给出了所有组的组名。现在如何获取每个组中的组成员列表..？

【问题讨论】：

标签： ios swift xmppframework

【解决方案1】：

首先，您必须创建一个房间并加入它，该房间已经存在，如下所示。

-(void)joinRoom:(NSString *)groupJid {

XMPPJID *JID = [XMPPJID jidWithString:groupJid];

_xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:_xmppRoomHS jid:JID];
[_xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
[_xmppRoom activate:_xmppStream];
[_xmppRoom joinRoomUsingNickname:_userNick history:nil];

[_xmppRoom fetchOwnersList];
[_xmppRoom fetchAdminsList];
[_xmppRoom fetchMembersList];
}

它将在其委托方法中获取所有成员、管理员和所有者列表

- (void)xmppRoom:(XMPPRoom *)sender didFetchOwnersList:(NSArray *)items {
}

- (void)xmppRoom:(XMPPRoom *)sender didFetchAdminsList:(NSArray *)items {
}

- (void)xmppRoom:(XMPPRoom *)sender didFetchMembersList:(NSArray *)items {
}

【讨论】：