Question

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"];

用户ABC@jabber.org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

在探索各种解决方案之后，我决定在这里编译和分享我的实现：

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];

检查此委托中是否已成功创建空间：
```
- (void)xmppRoomDidCreate:(XMPPRoom *)sender
```
检查您是否已加入此代表中的会议室：
```
- (void)xmppRoomDidJoin:(XMPPRoom *)sender
```

创建空间后，获取房间配置表单：

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

配置您的房间

/**
 * 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: Multi-User Chat，Implement Group Chat

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，您已经创建了一个XMPP多用户/组聊天室，并邀请了一位用户。：）

Answer 2

我觉得在alloc-init之后要做的第一件事是将它附加到你的xmppStream，所以它可以使用xmppStream来发送/接收消息。

更确切地说：

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"user101@conference.jabber.org/room" nickName:@"room"];
[room activate:[self xmppStream]];

//other things (create/config/...)

Answer 3

查看最新的XMPPMUCLight＆amp; XMPPRoomLight类似于Whatsapp和其他今天的趋势社交应用程序室，在线下或者没有人在室内时不会被摧毁或被踢。

请参阅此documentation＆amp; mod from MongooseIM

XMPPFramework - 如何创建MUC会议室并邀请用户？

3 个答案: