XMPPFramework IOS -实现

Question

提到这一点，我正在实现一个组聊天配置。

XMPPFramework - Implement Group Chat (MUC)

然而，作为参与者而不是版主，我无法获得成员名单。我试着读取多个堆栈答案，请求实现'muc#roomconfig_getmemberlist‘，但是XMPPRoom的获取配置委托在回调中没有给出这个字段值。

有谁能告诉我，实现这一点的确切方法是什么，以及如何获取成员列表。

Answer 1

默认情况下，这是在服务器中启用的配置，所以只需要设置，我们就必须自定义服务器，以使成员甚至离线和留出空间。因此，为了达到这个要求，像其他聊天应用程序成员一样要显示。

Answer 2

创建xmpp房间

/**
 This fuction is used to setup room with roomId
 */
-(void)setUpRoom:(NSString *)ChatRoomJID
{
    if (!ChatRoomJID)
    {
        return;
    }
    // Configure xmppRoom
    XMPPRoomMemoryStorage *roomMemoryStorage = [[XMPPRoomMemoryStorage alloc] init];

    XMPPJID *roomJID = [XMPPJID jidWithString:ChatRoomJID];

    xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomMemoryStorage jid:roomJID dispatchQueue:dispatch_get_main_queue()];

    [xmppRoom activate:xmppStream];
    [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];

    NSXMLElement *history = [NSXMLElement elementWithName:@"history"];
    [history addAttributeWithName:@" maxchars" stringValue:@"0"];
    [xmppRoom joinRoomUsingNickname:xmppStream.myJID.user
                            history:history
                           password:nil];


    [self performSelector:@selector(ConfigureNewRoom:) withObject:nil afterDelay:4];

}

/**
 This fuction is used configure new
 */
- (void)ConfigureNewRoom:(id)sender
{
    [xmppRoom configureRoomUsingOptions:nil];
    [xmppRoom fetchConfigurationForm];
    [xmppRoom fetchBanList];
    [xmppRoom fetchMembersList];
    [xmppRoom fetchModeratorsList];

}

创建房间后使用Xmpp房间的代理方法

- (void)xmppRoom:(XMPPRoom *)sender occupantDidJoin:(XMPPJID *)occupantJID withPresence:(XMPPPresence *)presence


- (void)xmppRoom:(XMPPRoom *)sender occupantDidLeave:(XMPPJID *)occupantJID withPresence:(XMPPPresence *)presence

使用这两种委托方法，您可以轻松地维护加入到MUC会议室的用户列表