Question

我正在开发一个利用Robie Hanson s XMPPFramework的SOSMPP聊天机。

最重要的功能已经落实——发送和接收信息。基本上,我已经建立了基本的实用聊天,当然很少见。

现在,我的问题是关于MUC。我从其他网站看到的代码显示,initWithRoomName 载于XMPPRoom。然而,这种方法在被遮盖的头巾中是不存在的。因此,除了此以外还有什么选择? 或者,如果没有,我如何利用XMPPFramework创建房间?

Thanks.

Answer 1

下面是我如何解决自己的问题。请注意,这一解决办法根本不涉及XMPPRoom。首先,我制定了一种方法,根据情况,可以创造或进入一个房间。 (Per XMPP文件,XML关于创建的要求与你寄送进入房间的要求相同;即,如果该房间在你进入该房间时还不存在,该服务将为你创建。)

我们来到这里。这是创造/进入房间的方法。这种方法是向您打算创建/进入的房间派出人员。如果你是第一个进入会议室,而该会议室尚未建立,你就自动成为房主和主持人。

- (void)createOrEnterRoom:(NSString *)roomName
{
//here we enter a room, or if the room does not yet exist, this method creates it
//per XMPP documentation: "If the room does not yet exist, the service SHOULD create the room"
//this method accepts an argument which is what you would baptize the room you wish created
NSXMLElement *presence = [NSXMLElement elementWithName:@"presence"];
NSString *room = [roomName stringByAppendingString:@"@conference.jabber.com/iMac"];
[presence addAttributeWithName:@"to" stringValue:room];
 NSXMLElement *x = [NSXMLElement elementWithName:@"x" xmlns:@"http://jabber.org/protocol/muc"];
NSXMLElement *history = [NSXMLElement elementWithName:@"history"];
 [history addAttributeWithName:@"maxstanzas" stringValue:@"50"];
 [x addChild:history];
 [presence addChild:x];
 [[self xmppStream] sendElement:presence];
}

其次,在宣布XMPPStream方法的Appelegate中,我们通过检查服务器发送的状态代码,以欺骗前方法对XML的反应进行过滤。如果地位法是201,本戈! 房间的设置只是罚款。除201条外,其他地位法意味着不同的事物,但为了我们的目的,应侧重于201条。

- (void)xmppStream:(XMPPStream *)sender didReceivePresence:(XMPPPresence *)presence
{
     NSXMLElement *x = [presence elementForName:@"x" xmlns:@"http://jabber.org/protocol/muc#user"];
    for (NSXMLElement *status in [x elementsForName:@"status"])
    {
        switch ([status attributeIntValueForName:@"code"])
        {
            case 201: [self notifyRoomCreationOk:room];
        }
    }
 }

Then, we tell the server that what we are creating a room of the type "instant," which means that we will send an IQ element telling it room defaults. notifyRoomCreationOk is a delegate method called in a different view when the room creation succeeds, after all I have to record the room in a text file to make it persistent so that the next time I open the app the room I created before will be visible. In my notifyRoomCreationOk method, I have sendDefaultRoomConfig which, basically, describes what is stated in the first sentence of this paragraph.

-(void)sendDefaultRoomConfig:(NSString *)room
{
NSXMLElement *x = [NSXMLElement elementWithName:@"x" xmlns:@"jabber:x:data"];
[x addAttributeWithName:@"type" stringValue:@"submit"];
NSXMLElement *query = [NSXMLElement elementWithName:@"query" xmlns:@"http://jabber.org/protocol/muc#owner"];
[query addChild:x];
XMPPIQ *iq = [XMPPIQ iq];
[iq addAttributeWithName:@"id" stringValue:[NSString stringWithFormat:@"inroom-cr%@", room]];
[iq addAttributeWithName:@"to" stringValue:room];
[iq addAttributeWithName:@"type" stringValue:@"set"];
[iq addChild:query];
[[self xmppStream ] sendElement:iq];
}

确保你有XMPPStream能够就所谓的上述方法发表意见,否则这些方法就赢得了工作。所有这一切都是如此。 Have!

Answer 2

    XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[email protected]/room" nickName:@"room"];
    [room createOrJoinRoom];
    [room sendInstantRoomConfig];
    [room setInvitedUser:@"[email protected]"];
    [room activate:[self xmppStream]];    
    [room inviteUser:jid1 withMessage:@"hello please join."];
    [room sendMessage:@"HELLO"];

the user [email protected] should receive the invite message

Answer 3

Your post is old, however now I would do it like this:

- (void)createRoomWithJid:(XMPPJID*)roomJID
{
    XMPPRoom *xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:self.xmppRoomHybridStorage
                                                           jid:roomJID
                                                 dispatchQueue:dispatch_get_main_queue()];
    [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
    [xmppRoom activate:self.xmppStream];

    [xmppRoom joinRoomUsingNickname:self.xmppStream.myJID.user
                            history:nil
                           password:nil];
}

Answer 4

采用XMPPFRAMWORK,按以下代码设立聊天室。

    let roomStorage: XMPPRoomMemoryStorage = XMPPRoomMemoryStorage()
    /**
     * Remember to add  conference  in your JID like this:
     * e.g. [email protected]
     */
    let roomJID: XMPPJID = XMPPJID.jidWithString("[email protected]")
    let xmppRoom: XMPPRoom = XMPPRoom(roomStorage: roomStorage,
        jid: roomJID,
        dispatchQueue: dispatch_get_main_queue())
    xmppRoom.activate(SKxmpp.manager().xmppStream)
    xmppRoom.addDelegate(self, delegateQueue: dispatch_get_main_queue())
    xmppRoom.joinRoomUsingNickname(SKxmpp.manager().xmppStream.myJID.user, history: nil, password: nil)
    xmppRoom.fetchConfigurationForm()

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