objective c - XMPP create group not getting Response in iOS -


i tried create new group using xmpp framework in ios using following code per given here

nsstring *nickname=[nsstring stringwithformat:@"%@.nickname@conference.server.hostname.in" ,newgroupname ];  xmpproommemorystorage * roommemory = [[xmpproommemorystorage alloc]init]; nsstring* roomid = [nsstring stringwithformat:@"%@@conference.server.hostname.in" ,newgroupname ]; xmppjid * roomjid = [xmppjid jidwithstring:roomid];  xmpproom* xmpproom = [[xmpproom alloc] initwithroomstorage:roommemory jid:roomjid dispatchqueue:dispatch_get_main_queue()];  [xmpproom activate:self.xmppstream]; [xmpproom adddelegate:self delegatequeue:dispatch_get_main_queue()]; [xmpproom joinroomusingnickname:nickname history:nil password:nil]; 

i did configure said in above specified thread with

[xmpproom fetchconfigurationform]; 

but there no response coming method

- (void)xmpproomdidcreate:(xmpproom *)sender{     ddlogverbose(@"%@: %@", this_file, this_method); } 

or not call coming method. means group not being created, right? no error shown or nothing in log.

please tell me mistake have made or if there more have this.

thanks in advance :d

do in way. in swift

    let roomstorage: xmpproommemorystorage = xmpproommemorystorage()     let roomjid: xmppjid = xmppjid.jidwithstring("chatroom10@conference.localhost")     let xmpproom: xmpproom = xmpproom(roomstorage: roomstorage,         jid: roomjid,         dispatchqueue: dispatch_get_main_queue())     xmpproom.activate(skxmpp.manager().xmppstream)     xmpproom.adddelegate(self, delegatequeue: dispatch_get_main_queue())     //xmpproom.configureroomusingoptions(nil)     xmpproom.joinroomusingnickname(skxmpp.manager().xmppstream.myjid.user, history: nil, password: nil)     xmpproom.fetchconfigurationform() 

Comments

Popular posts from this blog

Email notification in google apps script -

c++ - Difference between pre and post decrement in recursive function argument -

javascript - IE11 incompatibility with jQuery's 'readonly'? -