XMPPFramework - 如何创建MUC会议室并邀请用户?

Nav*_*afi 18 xmpp objective-c ios xmppframework

我正在使用Robbiehanson的iOS XMPPFramework.我正在尝试创建一个MUC房间并邀请用户进入群聊室,但它无法正常工作.

我使用以下代码:

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"user101@conference.jabber.org/room" nickName:@"room"];
[room createOrJoinRoom];
[room sendInstantRoomConfig];
[room setInvitedUser:@"ABC@jabber.org"];
[room activate:[self xmppStream]];    
[room inviteUser:jid1 withMessage:@"hello please join."];
[room sendMessage:@"HELLO"];
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用户ABC@jabber.org应该收到邀请消息但没有发生任何事情.

任何帮助将不胜感激.:)

Kei*_*OYS 33

在探索了各种解决方案之后,我决定在这里编译和共享我的实现:

  1. 创建XMPP会议室:

    XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];
    
    /** 
     * Remember to add 'conference' in your JID like this:
     * e.g. uniqueRoomJID@conference.yourserverdomain
     */
    
    XMPPJID *roomJID = [XMPPJID jidWithString:@"chat@conference.shakespeare"];
    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];
    
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  2. 检查是否在此委托中成功创建了房间:

    - (void)xmppRoomDidCreate:(XMPPRoom *)sender
    
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  3. 检查您是否已加入此代表中的房间:

    - (void)xmppRoomDidJoin:(XMPPRoom *)sender
    
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  4. 创建房间后,获取房间配置表格:

    - (void)xmppRoomDidJoin:(XMPPRoom *)sender {
        [sender fetchConfigurationForm];
    }
    
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  5. 配置你的房间

    /**
     * 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];
    }
    
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    参考文献:XEP-0045:多用户聊天,实施群聊

  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!"];
    }
    
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在那里,您已经创建了一个XMPP多用户/组聊天室,并邀请了一位用户.:)

  • @rohitmandiwal我很高兴!你可以通过这条线创建一个受密码保护的MUC房间,如上所示 - `[xmppRoom joinRoomUsingNickname:[self appDelegate] .xmppStream.myJID.user history:nil password:@"myPassword"];` (2认同)