Open NileshTripathi opened 10 years ago
Find stackoverflow link for proper visibility: http://stackoverflow.com/questions/22252257/cant-fetch-detail-of-xmpproom-from-the-server-xmpp-framework
And google groups link: https://groups.google.com/forum/#!topic/xmppframework/oKNwXOQvz5A
Thanks in advance.
Hey robbiehanson,
I have already posted this question on google grouops, stackoverflow and then I came here with some hope that some one will help me.
I have already did enough R&D and then I am here to question something.
Please find below brief detail of the question
I have created a MembersOnly(non Public), Persistant Room. In which I have invited multiple occupants. Now I want to fetch detail of room for the user who were invited by the creator.
I tried this code :
XMPP requirement :
<iq from='hag66@shakespeare.lit/pda' id='ik3vs715' to='coven@chat.shakespeare.lit' type='get'>
iOS code on view will apear: NSXMLElement *query = [NSXMLElement elementWithName:@"query" xmlns:@"http://jabber.org/protocol/disco#info"];//
NSString *iqID = [[appDelegate xmppStream] generateUUID];
XMPPJID jID = self.room.roomJID; XMPPIQ element = [XMPPIQ iqWithType:@"get" to:jID elementID:iqID child:query]; [element addAttributeWithName:@"from" stringValue:[[[appDelegate xmppStream] myJID] full]]; [[appDelegate xmppStream] fetchInformationForGivenIQ:element];
This should provide me this kinda result, which is I am getting public rooms I have created but not for other non public rooms:
<iq from='coven@chat.shakespeare.lit' id='ik3vs715' to='hag66@shakespeare.lit/pda' type='result'>