OGeek|极客世界-中国程序员成长平台

标题: ios - 无法在 iphone 中使用 Xmpp 在群聊中发送和接收消息 [打印本页]

---

作者: 菜鸟教程小白    时间: 2022-12-12 10:57
标题: ios - 无法在 iphone 中使用 Xmpp 在群聊中发送和接收消息

使用 xmpp 我可以创建群组并向 friend 发送邀请，但是当我在群组上发送消息时，成员将永远不会收到该消息。

成员(member)必须接受邀请吗？如果是，请告诉我怎么做？

请引用下面的代码，如果我犯了任何错误或仍然遗漏任何东西，请指导我，以便在我可以在群组中发送和接收消息并与 friend 聊天之后。

下面我附上了一些代码片段，用于在 xmpp 中创建组并发送消息。

[self setUpRoom:[NSString stringWithFormat"%@@conference.myserver",@"GroupName"]];


-(void)setUpRoomNSString *)ChatRoomJID {

    // Configure xmppRoom
    XMPPRoomMemoryStorage *roomMemoryStorage = [[XMPPRoomMemoryStorage alloc] init];

    XMPPJID *roomJID = [XMPPJID jidWithString:ChatRoomJID];

    xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomMemoryStorage jid:roomJID dispatchQueue:dispatch_get_main_queue()];

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


    [xmppRoom joinRoomUsingNickname:_ro(@"LoginNumber")
                            history:nil
                           password:nil];

    [self performSelectorselector(ConfigureNewRoom withObject:nil afterDelay:4];

}

现在房间确认我使用了这个片段

- (void)ConfigureNewRoomid)sender
{
    [xmppRoom configureRoomUsingOptions:nil];
    [xmppRoom fetchConfigurationForm];
    [xmppRoom fetchBanList];
    [xmppRoom fetchMembersList];
    [xmppRoom fetchModeratorsList];

}

XMPP Room 委托(delegate)方法

- (void)xmppRoomDidCreateXMPPRoom *)sender
{
    DDLogInfo(@"%@: %@", THIS_FILE, THIS_METHOD);

   // I am inviting friends after room is created

    for (int i = 0; i<[self.friendListArray count]; i++)
    {
        NSString * tempStr=[NSString stringWithFormat"%@@myserver",[[self.friendListArray objectAtIndex:i] valueForKey"UserNumber"]];
        [sender inviteUser:[XMPPJID jidWithString:tempStr] withMessage"Greetings!"];
    }

}

- (void)xmppRoomDidJoinXMPPRoom *)sender
  {
        DDLogInfo(@"%@: %@", THIS_FILE, THIS_METHOD);

        NSLog(@"........Room Did join.......");
  }

- (void)xmppRoomXMPPRoom *)sender didFetchConfigurationFormNSXMLElement *)configForm
   {
        DDLogInfo(@"%@: %@", THIS_FILE, THIS_METHOD);

        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"]];
            }

            if ([var isEqualToString:@"roomconfig_enablelogging"])
            {

                [field removeChildAtIndex:0];
                [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]];
            }

            if ([var isEqualToString:@"muc#roomconfig_maxusers"])
            {

                [field removeChildAtIndex:0];
                [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"100"]];
            }


        }
        [sender configureRoomUsingOptions:newConfig];
  }

在按钮点击时分组发送消息以进行测试

-(void)sendGroupMessage
{
    [xmppRoom sendMessageWithBody:@"Hi All"];

    NSXMLElement *x = [NSXMLElement elementWithName:@"groupchat" xmlns:XMPPMUCNamespace];

    XMPPMessage *message = [XMPPMessage message];
    [message addAttributeWithName:@"to" stringValue:[NSString stringWithFormat:@"%@/%@",[xmppRoom.roomJID full],_ro(@"LoginNumber")]];
    [message addChild:x];
    NSLog(@"x in Invite === %@",x);
    [xmppStream sendElement:message];
}

---

Best Answer-推荐答案

代替:

 [xmppStream sendElement:message]

尝试:

 [xmppRoom sendMessage:message]

关于ios - 无法在 iphone 中使用 Xmpp 在群聊中发送和接收消息，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24993186/

---

欢迎光临 OGeek|极客世界-中国程序员成长平台 (http://ogeek.cn/)

Powered by Discuz! X3.4