ios - XMPPFramework - 如何创建 MUC 房间并邀请用户？

Question

我正在使用 Robbiehanson 的 iOS XMPPFramework。我正在尝试创建一个 MUC 房间并邀请用户加入群聊室，但它不起作用。

我正在使用以下代码：

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 应该会收到邀请消息，但没有任何反应。

任何帮助将不胜感激。:)

score 35 · Accepted Answer

在探索了各种解决方案之后，我决定在这里编译并分享我的实现：

创建一个 XMPP 房间：

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];

检查此委托中是否成功创建房间：

- (void)xmppRoomDidCreate:(XMPPRoom *)sender

检查您是否已加入此代表的房间：
```
- (void)xmppRoomDidJoin:(XMPPRoom *)sender
```

房间创建后，获取房间配置表：

- (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：XEP-0045：多用户聊天，实现群聊

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，您创建了一个 XMPP 多用户/群组聊天室，并邀请了一个用户。:)

score 1 · Accepted Answer

我有一种感觉，在 alloc-init 之后要做的第一件事就是将它附加到您的 xmppStream 中，这样它就可以使用 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/...)

score 1 · Accepted Answer

查看最新的 XMPPMUCLight 和 XMPPRoomLight，它类似于 Whatsapp 和其他当今趋势的社交应用程序房间，这些应用程序房间在离线或房间内无人时不会被破坏或成员被踢。

请参阅MongooseIM 的此文档和mod

ios - XMPPFramework - 如何创建 MUC 房间并邀请用户？

3 回答 3

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