[英]Add a user to xmpp group or in muc
我想將4個成員(已經是oprefire中的用戶)添加到組中。 我想在沒有用戶許可或發送邀請的情況下添加它們
現在,我邀請用戶使用此代碼:
[sender inviteUser:[XMPPJID jidWithString:@"keithoys"] withMessage:@"Greetings!"];
還有其他方法可以做到這一點嗎?
當用戶向其他用戶發送請求時,將調用以下委托方法:
-(void)xmppMUC:(XMPPMUC *)sender roomJID:(XMPPJID *)roomJID didReceiveInvitation:(XMPPMessage *)message{
roomMemoryStorage = [[XMPPRoomMemoryStorage alloc] init];
xmppRoom = [[XMPPRoom alloc]
initWithRoomStorage:roomMemoryStorage
jid:roomJID
dispatchQueue:dispatch_get_main_queue()];
[xmppRoom activate:[self xmppStream]];
[xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
//Now add user to the group directly without prompting them
[xmppRoom joinRoomUsingNickname:[xmppStream myJID].user history:nil];
}
我只寫了代碼,如果您需要解釋,我會的。
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