在我的聊天应用程序中,我想获得所有在线注册用户。所以大家不仅人在我的名册随同此代码实现:iOS XMPP框架获取所有注册用户
- (void)xmppStream:(XMPPStream *)sender didReceivePresence:(XMPPPresence *)presence {
// a buddy went offline/online
NSString *presenceType = [presence type]; // online/offline
NSString *myUsername = [[sender myJID] user];
NSString *presenceFromUser = [[presence from] user];
if (![presenceFromUser isEqualToString:myUsername]) {
if ([presenceType isEqualToString:@"available"]) {
[_chatDelegate newBuddyOnline:[NSString stringWithFormat:@"%@@%@", presenceFromUser, @"chat.denederlandsewateren.nl"]];
} else if ([presenceType isEqualToString:@"unavailable"]) {
[_chatDelegate buddyWentOffline:[NSString stringWithFormat:@"%@@%@", presenceFromUser, @"chat.denederlandsewateren.nl"]];
}
}
}
这段代码的用户只能看到谁是“朋友”,但我需要在这个特定域的所有注册用户的其他用户。这可能与ejabberd?
“DOMAIN”是组的JID?你能上传代码吗?我无法获得该组成员的名单。 – Apple
嗨@Mark Molina,你能帮我找到所有的用户吗?我坚持这个问题,上面的答案不适合我。 – brainforked
'queryElements'在'didReceiveIQ:' –