Question

我可以使用XMPPFramework创建MUC，并使用以下代码发送用户邀请请求加入该会议室。

// Creating
AppDelegate *dele =(AppDelegate *) [[UIApplication sharedApplication]delegate];

xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:dele jid:[XMPPJID jidWithString:self.roomName] dispatchQueue:dispatch_get_main_queue()];
[xmppRoom addDelegate:dele delegateQueue:dispatch_get_main_queue()];
[xmppRoom activate:dele.xmppStream];
[xmppRoom joinRoomUsingNickname:self.myNick history:nil];

// Inviting
[xmppRoom inviteUser:[XMPPJID jidWithString:@"abc@host"] withMessage:@"Come Join me"];

用户“abc”如何知道他已收到邀请，他如何通过接受或拒绝对此做出反应？

我在XMPPFramework中找不到任何直接处理聊天室邀请的课程。我的研究表明，每当用户收到聊天室邀请时，都会调用xmmppStream的委托方法：

- (void)xmppStream:(XMPPStream *)sender didReceiveMessage:(XMPPMessage *)message

在该消息中，我检查它是否包含名为“invite”的NSXMLElement，如果它包含，则我向用户发送回调。然后我创建聊天室，其名称与用户收到邀请的聊天室名称相同，并进入新创建的房间。它工作正常，但安静冗长，不安静有效。我想知道XMPPFramework中是否有可用的here类可以单独处理聊天室邀请。例如，检测，接受和拒绝房间邀请。

我的提取房间名称的代码：

- (void)xmppStream:(XMPPStream *)sender didReceiveMessage:(XMPPMessage *)message
{
    NSXMLElement * x = [message elementForName:@"x" xmlns:XMPPMUCUserNamespace];
    NSXMLElement * invite  = [x elementForName:@"invite"];
    NSXMLElement * decline = [x elementForName:@"decline"];
    NSXMLElement * directInvite = [message elementForName:@"x" xmlns:@"jabber:x:conference"];
    NSString *msg = [[message elementForName:@"body"]stringValue];
    NSString *from = [[[message attributeForName:@"from"]stringValue];
    if (invite || directInvite)
    {
        [self createAndEnterRoom:from Message:msg];
        return;
    }
    [self.delegate newMessageRecieved:msg];
}

Answer 1

有关会议室邀请和拒绝，请实施XMPPMUCDelegate及其方法-xmppMUC:didReceiveRoomInvitation:和-xmppMUC:didReceiveRoomInvitationDecline:。

要获得房间JID，请调用[message from];

要加入会议室，请实例化XMPPRoom并调用-joinRoomUsingNickname:history:。

然后让你的房间委托类实现XMPPRoomDelegate，并实现一些delegate methods来处理房间中的接收消息。

看起来目前还没有更自动的方式来回复邀请。

更新：委托回调现在接收房间JID作为参数，稍微澄清了语义。

- (void)xmppMUC:(XMPPMUC *)sender roomJID:(XMPPJID *) roomJID didReceiveInvitation:(XMPPMessage *)message;
- (void)xmppMUC:(XMPPMUC *)sender roomJID:(XMPPJID *) roomJID didReceiveInvitationDecline:(XMPPMessage *)message;

Answer 2

只需添加以下代码

if  ([presenceType isEqualToString:@"subscribe"]) {

     [_chatDelegate newBuddyOnline:[NSString stringWithFormat:@"%@@%@", presenceFromUser, @"localhost"]];
     NSLog(@"presence user wants to subscribe %@",presenceFromUser);

     [xmppRoster acceptPresenceSubscriptionRequestFrom:[presence from] andAddToRoster:YES];

 //For reject button
//     [xmppRoster rejectPresenceSubscriptionRequestFrom:[tmpPresence from]];          
}

方法内的

 - (void)xmppStream:(XMPPStream *)sender didReceivePresence:(XMPPPresence *)presence ;
method

接受聊天室邀请

2 个答案: