iOS XMPP framework get all registered users - objective-c

Question

In my chat application I want to get all online registered users. So everybody and not only people in my roster which is achieved with this code:
- (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"]];
}
}
}
With this code the users only sees other users who are 'friends' but I need all users registered on this particular domain. Is this possible with ejabberd?

Answer 1

- (void)getAllRegisteredUsers {
NSError *error = [[NSError alloc] init];
NSXMLElement *query = [[NSXMLElement alloc] initWithXMLString:#"<query xmlns='http://jabber.org/protocol/disco#items' node='all users'/>"
error:&error];
XMPPIQ *iq = [XMPPIQ iqWithType:#"get"
to:[XMPPJID jidWithString:#"DOMAIN"]
elementID:[xmppStream generateUUID] child:query];
[xmppStream sendElement:iq];
}
- (BOOL)xmppStream:(XMPPStream *)sender didReceiveIQ:(XMPPIQ *)iq
{
NSXMLElement *queryElement = [iq elementForName: #"query" xmlns: #"http://jabber.org/protocol/disco#items"];
if (queryElement) {
NSArray *itemElements = [queryElement elementsForName: #"item"];
NSMutableArray *mArray = [[NSMutableArray alloc] init];
for (int i=0; i<[itemElements count]; i++) {
NSString *jid=[[[itemElements objectAtIndex:i] attributeForName:#"jid"] stringValue];
[mArray addObject:jid];
}
}

Answer 2

I had the same issue, I got queryElement as nil as well. I've changed the response code to see the XML like this:
- (BOOL)xmppStream:(XMPPStream *)sender didReceiveIQ:(XMPPIQ *)iq
{
//DDLogVerbose(#"%#: %# - %#", THIS_FILE, THIS_METHOD, [iq elementID]);
//NSXMLElement *queryElement = [iq elementForName:#"query" xmlns: #"http://jabber.org/protocol/disco#items"];
NSXMLElement *queryElement = [iq elementForName:#"query" xmlns: #"jabber:iq:roster"];
NSLog(#"IQ: %#",iq);
if (queryElement) {
NSArray *itemElements = [queryElement elementsForName: #"item"];
NSMutableArray *mArray = [[NSMutableArray alloc] init];
for (int i=0; i<[itemElements count]; i++) {
NSString *jid=[[[itemElements objectAtIndex:i] attributeForName:#"jid"] stringValue];
NSLog(#"%#",jid);
[mArray addObject:jid];
}
}
return NO;
}
As you may see what I've changed is the xmlns: from this xmlns: #"http://jabber.org/protocol/disco#items" to this xmlns: #"jabber:iq:roster" and that gave me the list of users.
I'm using ejabberd, not sure if this works for all the others XMPP servers.
Also I've found that this gave me the list of the "buddy" users, looks like if you want "all" users you need to make the query as an admin user. Please check this link for more information about it: https://www.ejabberd.im/node/3420

Answer 3

After googling, You can not get all user easily, You must need to create Shared Roster Groups by follow step in the Example 1: everybody can see everybody else after done this you will get all the online users in the below delegate methods.
- (void)xmppStream:(XMPPStream *)sender didReceivePresence:(XMPPPresence *)presence