XMPPFramework - 如何创建MUC房间并邀请用户？

发布于 2024-11-25 13:37:23 字数 887 浏览 3 评论 0原文

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房间，但它不起作用。

我使用以下代码：

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

用户 [email protected] 应该收到邀请消息但什么也没有发生。

任何帮助将不胜感激。 :)

原文

I am using Robbiehanson's iOS XMPPFramework. I am trying to create a MUC room and invite a user to the group chat room but it is not working.

I am using the following code:

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

The user [email protected] should receive the invite message but nothing is happening.

Any help will be appreciated. :)

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生寂 2024-12-02 13:37:23

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 XMPP Room：

XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];

/** 
 * 请记住在您的 JID 中添加“会议”，如下所示：
 * 例如 [电子邮件受保护]
 */

XMPPJID *roomJID = [XMPPJID jidWithString:@"[电子邮件受保护]"] ;
XMPPRoom *xmppRoom = [[XMPPRoom 分配] initWithRoomStorage:roomStorage
                                                       jid:房间JID
                                             调度队列:dispatch_get_main_queue()];

[xmppRoom activate:[self appDelegate].xmppStream];
[xmppRoom addDelegate:self 
        delegateQueue:dispatch_get_main_queue()];

[xmppRoom joinRoomUsingNickname:[self appDelegate].xmppStream.myJID.user 
                        历史：无 
                       密码：无]；

检查该委托中是否成功创建房间：

- (void)xmppRoomDidCreate:(XMPPRoom *)发送者

检查您是否已加入此委托中的房间：
```
- (void)xmppRoomDidJoin:(XMPPRoom *)发送者
```

创建房间后，获取房间配置表单：

- (void)xmppRoomDidJoin:(XMPPRoom *)sender {
    [发件人获取配置表单]；
}

配置你的房间
<前><代码>/**
* 防止此消息所必需的：
*“在确认配置之前，该房间将被锁定，无法进入。”
*/
- (void)xmppRoom:(XMPPRoom *)发送者 didFetchConfigurationForm:(NSXMLElement *)configForm
{
NSXMLElement *newConfig = [configForm 副本];
NSArray *fields = [newConfig elementsForName:@"field"];
for (NSXMLElement *字段中的字段)
{
NSString *var = [字段 attributeStringValueForName:@"var"];
// 使空间持久化
if ([var isEqualToString:@"muc#roomconfig_persistentroom"]) {
[字段removeChildAtIndex:0];
[字段 addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]];
}
}
[发送者configureRoomUsingOptions:newConfig];
}
参考文献：XEP-0045：多用户聊天，< a href="https://stackoverflow.com/questions/19268629/xmpp-ios-chat-client-implement-group-chat">实施组聊天

邀请用户

- (void)xmppRoomDidJoin:(XMPPRoom *)发送者 
{
    /** 
     * 您可以从包含 for 循环参与者的数组中读取 
     * 并在此处以相同方式发送多个邀请
     */

    [发件人邀请用户：[XMPPJID jidWithString：@“keithoys”] withMessage：@“问候！”];
}

在这里，您创建了一个 XMPP 多用户/群组聊天室，并邀请了一位用户。 :)

After exploring various solutions, I've decided to compile and share my implementation here:

Create an XMPP Room:

XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];

/** 
 * Remember to add 'conference' in your JID like this:
 * e.g. [email protected]
 */

XMPPJID *roomJID = [XMPPJID jidWithString:@"[email protected]"];
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];

Check if room is successfully created in this delegate:
```
- (void)xmppRoomDidCreate:(XMPPRoom *)sender
```
Check if you've joined the room in this delegate:
```
- (void)xmppRoomDidJoin:(XMPPRoom *)sender
```

After room is created, fetch room configuration form:

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

Configure your room

/**
 * 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];
}

References: XEP-0045: Multi-User Chat, Implement Group Chat

Invite users

- (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!"];
}

There, you've created a XMPP multi-user/group chat room, and invited a user. :)

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枕头说它不想醒 2024-12-02 13:37:23

我感觉alloc-init之后要做的第一件事就是将它附加到你的xmppStream，这样它就可以使用xmppStream发送/接收消息。

更准确地说：

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[email protected]/room" nickName:@"room"];
[room activate:[self xmppStream]];

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

I have the feeling that the first thing to do after alloc-init is to attach it to your xmppStream, so it can use xmppStream to send/receive messages.

More exactly:

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[email protected]/room" nickName:@"room"];
[room activate:[self xmppStream]];

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

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