XMPPFramework - 如何创建一个多用户聊天室并邀请其他用户？

Question

XMPPFramework - 如何创建一个多用户聊天室并邀请其他用户？

19

我正在使用Robbiehanson的iOS XMPPFramework。我尝试创建一个MUC房间并邀请用户加入群聊，但没有成功。

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应该收到邀请消息，但没有任何反应。

非常感谢您的帮助。:)

- Naveed Rafi

3个回答

1

我有一种感觉，在分配初始化之后，第一件事就是将其附加到您的xmppStream上，以便它可以使用xmppStream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/...)

- Codrut

1

检查最新的XMPPMUCLight和XMPPRoomLight，它们类似于WhatsApp和其他当今流行的社交应用程序房间，即使离线或没有人在房间中，也不会被销毁或成员踢出。

请参考此文档和MongooseIM的mod。

- codelover

你能帮我解决这个问题吗？https://stackoverflow.com/questions/44172852/how-to-parse-xmppmessage-element-attribute-and-node-in-xmppframework-with-swif ? - May Phyu

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- Keith OYS · Accepted Answer

在探索了各种解决方案后，我决定在此处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];

检查此代理中的房间是否成功创建：
```
- (void)xmppRoomDidCreate:(XMPPRoom *)sender
```
检查您是否在此委托中加入了房间：

- (void)xmppRoomDidJoin:(XMPPRoom *)sender

创建房间后，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];
}

参考资料： XEP-0045：多用户聊天，实现群组聊天

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!"];
}