Question

我正在使用git上提供的Robby Hanson的XMPP库，我正在尝试实现MUC或群聊室。

我正在使用一个用户创建房间，然后尝试加入，而不邀请其他用户。问题是，如果我尝试与另一个用户连接，而不是房间的创建者，我会收到错误：

<iq xmlns="jabber:client" type="error" id="A7F05488-4A84-4EC0-8A6C-0F1541690534" from="newroom4@conference.administrator" to="newuser229@administrator/abdbd1bc"><query xmlns="http://jabber.org/protocol/muc#admin"><item affiliation="member"/></query><error code="403" type="auth"><forbidden xmlns="urn:ietf:params:xml:ns:xmpp-stanzas"/></error></iq>

此外，我搜索了错误，发现如果用户被禁止，可能会出现错误403。这不是这种情况。所以当我尝试获取像fetchConfigurationForm或fetchMembersList这样的房间信息时会发生错误。

所以，这是我正在使用的代码：

- (void)testGroupButtonFunction{
    XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];
    XMPPJID *roomJID = [XMPPJID jidWithString:@"newRoom4@conference.administrator"];

    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];
}

- (void)handleDidJoinRoom:(XMPPRoom *)room withNickname:(NSString *)nickname{

    NSLog(@"handleDidJoinRoom");

}

- (void)handleIncomingMessage:(XMPPMessage *)message room:(XMPPRoom *)room{

    NSLog(@"Incomming message: %@", message.debugDescription);

}

- (void)handleOutgoingMessage:(XMPPMessage *)message room:(XMPPRoom *)room{

    NSLog(@"Outgoing message: %@", message.debugDescription);

}

- (void)xmppRoom:(XMPPRoom *)sender didFetchMembersList:(NSArray *)items{

    NSLog(@"didFetchMembersList: %@", items.debugDescription);

}

- (void)xmppRoom:(XMPPRoom *)sender didNotFetchMembersList:(XMPPIQ *)iqError{

    NSLog(@"didNotFetchMembersList error: %@", iqError.debugDescription);

}

- (void)xmppRoomDidCreate:(XMPPRoom *)sender{

    NSLog(@"xmppRoomDidCreate");

}

- (void)xmppRoom:(XMPPRoom *)sender didConfigure:(XMPPIQ *)iqResult{

    NSLog(@"didConfigure: %@", iqResult.debugDescription);

}

- (void)xmppRoomDidJoin:(XMPPRoom *)sender {

    NSLog(@"xmppRoomDidJoin");
// I use the same code to create or join a room that's why I commented the next line
//    [xmppRoom fetchConfigurationForm];
    //Next line generates the error:
    [xmppRoom fetchMembersList];

}

- (void)xmppRoom:(XMPPRoom *)sender didFetchConfigurationForm:(NSXMLElement *)configForm{

    NSLog(@"didFetchConfigurationForm");

    NSXMLElement *newConfig = [configForm copy];
    NSArray *fields = [newConfig elementsForName:@"field"];
    for (NSXMLElement *field in fields)
    {
        NSString *var = [field attributeStringValueForName:@"var"];
        NSLog(@"didFetchConfigurationForm: %@", var);
        // Make Room Persistent
        if ([var isEqualToString:@"muc#roomconfig_persistentroom"]) {
            [field removeChildAtIndex:0];
            [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]];
        }
        if ([var isEqualToString:@"muc#roomconfig_roomdesc"]) {
            [field removeChildAtIndex:0];
            [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"Apple"]];
        }
    }

    [sender configureRoomUsingOptions:newConfig];

}

- (void)xmppRoom:(XMPPRoom *)sender didNotConfigure:(XMPPIQ *)iqResult{

    NSLog(@"didNotConfigure: %@",iqResult.debugDescription);

}

我使用相同的代码创建或加入一个房间，这就是我评论下一行的原因：

[xmppRoom fetchConfigurationForm];

另外我想添加我设置：

publicRoom：1 审核：0 membersOnly：0 canInvite：1 roomPassword：无 canRegister：1 canDiscoverJID：1 logEnabled：1

此外，如果我尝试从一个设备发送消息，当我在另一个用户（该用户不是该组的创建者/管理员）记录的第二个设备上检索消息时，我看到该消息在控制台使用LOG_LEVEL_VERBOSE，但它没有调用委托方法。知道为什么没有调用委托方法？（我确实在.h中添加了XMPPRoomDelegate）任何人都可以帮我解决这个错误吗？非常感谢您的耐心和支持！

Answer 1

是因为它不遵循XEP0045指令。你应该做一个实现这个方法的类别：

- (void)joinRoomByContact:(Contact *)contact history:(NSXMLElement *)history
{
    dispatch_block_t block = ^{ @autoreleasepool {

        // Check state and update variables

        // <presence to='darkcave@chat.shakespeare.lit/firstwitch'>
        //   <x xmlns='http://jabber.org/protocol/muc'/>
        //     <history/>
        //     <password>passwd</password>
        //   </x>
        // </presence>

        NSXMLElement *x = [NSXMLElement elementWithName:@"x" xmlns:XMPPMUCNamespace];
        if (history)
        {
            [x addChild:history];
        }

        //XMPPPresence *presence = [XMPPPresence presenceWithType:nil to:myRoomJID];
        XMPPElement * presence = [[XMPPElement alloc] initWithName:@"presence"];
        [presence addAttributeWithName:@"to" stringValue:[NSString stringWithFormat:@"%@/%@",roomJID.bare,contact.name]];
        [presence addAttributeWithName:@"from" stringValue:contact.jid];
        [presence addChild:x];


        [xmppStream sendElement:presence];


        state |= (1 << 3);

    }};

    if (dispatch_get_specific(moduleQueueTag))
        block();
    else
        dispatch_async(moduleQueue, block);
}

我希望能帮到你;）

如何使用iOS XMPP Robbie Hanson库获取具有成员的MUC会员的成员列表

1 个答案: