我正在使用 git 上可用的 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，而不邀请另一个用户。问题是，如果我尝试连接另一个用户，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>
我还搜索了错误，发现如果用户被禁止，可能会出现错误 403。这不是这里的情况。因此，当我尝试获取诸如 fetchConfigurationForm 或 fetchMembersList 之类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);
}
我使用相同的代码来创建或加入房间，这就是我评论下一行的原因：
[xmppRoom fetchConfigurationForm];
另外我想补充一下我设置的：
publicRoom : 1 主持 : 0 membersOnly : 0 canInvite : 1 roomPassword : nil canRegister : 1 canDiscoverJID : 1 logEnabled : 1
此外，如果我尝试从一个设备发送消息，当我在与另一个用户（该用户不是组的创建者/管理员）记录的第二个设备上检索消息时，我会使用 LOG_LEVEL_VERBOSE 在控制台中看到传入消息，但它不调用委托方法。知道为什么不调用委托方法吗？（我确实在 .h 中添加了 XMPPRoomDelegate）有人可以帮我解决这个错误吗？非常感谢您的耐心和支持！