本文介绍了Xmpp IOS多用户聊天。我没有找到接受小组邀请的方法?我怎么能接受这个邀请函的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我发送邀请时,此功能被调用,但我无法理解代码行应该用于接受邀请* 。我试图创建一个多用户和多组邀请也
称为收到消息功能。

When i send invitation this function is called but i can't understand what line of code should use for accept invitation*. am trying to create a multi user and multi groups invitation alsocalled did received message function.

- (void)xmppMUC:(XMPPMUC *) sender roomJID:(XMPPJID *) roomJID didReceiveInvitation:(XMPPMessage *)message
{
}


推荐答案

接受来电邀请:

- (void)xmppMUC:(XMPPMUC *)sender roomJID:(XMPPJID *) roomJID didReceiveInvitation:(XMPPMessage *)message
{ XMPPRoom *mu = [[XMPPRoom alloc] initWithRoomStorage:xmpproomMstorage jid:roomJID
                                           dispatchQueue:dispatch_get_main_queue()];

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

    self.toSomeOne = roomJID;

    [mu activate: self.xmppStream];
    [mu fetchConfigurationForm];
    [mu addDelegate:self delegateQueue:dispatch_get_main_queue()];
    [mu joinRoomUsingNickname:xmppStream.YourJid.user history:nil password:@"Your Password"];
self.toSomeOne = roomJID;
    XMPPPresence *presence = [XMPPPresence presence];
   [[self xmppStream] sendElement:presence];
    [xmppRoster addUser:roomJID  withNickname:roomJID.full];
    [self goOnline];
}

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09-09 12:04