XMPPFramework – 如何创buildMUC房间并邀请用户？

我正在使用Robbiehanson的iOS XMPPFramework。 我正在尝试创build一个MUC房间，并邀请用户join群组聊天室，但这不起作用。

我正在使用下面的代码：

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

用户[e-mail]应该收到邀请信息，但没有任何事情发生。

任何帮助将不胜感激。 🙂

Answer 1

在探索了各种解决scheme后，我决定在这里编译和分享我的实现：

1. 创build一个XMPP房间：

    XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init]; /** * Remember to add 'conference' in your JID like this: * eg [e-mail] */ XMPPJID *roomJID = [XMPPJID jidWithString:@"[e-mail]"]; XMPPRoom *xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomStorage jid:roomJID dispatchQueue:dispatch_get_main_queue()]; [xmppRoom activate:[self appDelegate].xmppStream]; [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()]; [xmppRoom joinRoomUsingNickname:[self appDelegate].xmppStream.myJID.user history:nil password:nil]; 

2. 检查此代表是否成功创build了会议室：

    - (void)xmppRoomDidCreate:(XMPPRoom *)sender 

3. 检查你是否join了这个代表的房间：

    - (void)xmppRoomDidJoin:(XMPPRoom *)sender 

4. 创build房间后，取房间configuration表：

    - (void)xmppRoomDidJoin:(XMPPRoom *)sender { [sender fetchConfigurationForm]; } 

5. configuration你的房间

    /** * Necessary to prevent this message: * "This room is locked from entry until configuration is confirmed." */ - (void)xmppRoom:(XMPPRoom *)sender didFetchConfigurationForm:(NSXMLElement *)configForm { NSXMLElement *newConfig = [configForm copy]; NSArray *fields = [newConfig elementsForName:@"field"]; for (NSXMLElement *field in fields) { NSString *var = [field attributeStringValueForName:@"var"]; // Make Room Persistent if ([var isEqualToString:@"muc#roomconfig_persistentroom"]) { [field removeChildAtIndex:0]; [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]]; } } [sender configureRoomUsingOptions:newConfig]; } 

   参考文献： XEP-0045：多用户聊天 ， 实现群聊

6. 邀请用户

    - (void)xmppRoomDidJoin:(XMPPRoom *)sender { /** * You can read from an array containing participants in a for-loop * and send multiple invites in the same way here */ [sender inviteUser:[XMPPJID jidWithString:@"keithoys"] withMessage:@"Greetings!"]; } 

在那里，你已经创build了一个XMPP多用户/组聊天室，并邀请了一个用户。 🙂

Answer 2

我有这样的感觉，alloc-init之后要做的第一件事就是将它附加到你的xmppStream，所以它可以使用xmppStream发送/接收消息。

更确切地说：

 XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[e-mail]/room" nickName:@"room"]; [room activate:[self xmppStream]]; //other things (create/config/...) 

Answer 3

查看最新的XMPPMUCLight＆XMPPRoomLight，它类似于Whatsapp和其他今天的趋势社交应用程序的房间，不会被破坏或成员在离线或室内没有人踢。

从MongooseIM中引用这个文档和mod