multiuserchat

how to create persistent muc room in smack 4.1 beta2

痞子三分冷 提交于 2019-12-03 18:22:06
migrated from asmack to smack 4.1 beta2. The muc rooms created are no longer persistent. MultiUserChatManager mucm=MultiUserChatManager.getInstanceFor(connection); muc=mucm.getMultiUserChat(groupid+"@conference.localhost"); DiscussionHistory histroy=new DiscussionHistory(); histroy.setMaxStanzas(10); muc.createOrJoin(username,null,histroy,SmackConfiguration.getDefaultPacketReplyTimeout()); muc.nextMessage(); when created with gajim, the rooms are persistent. EDIT : Here is code we used earlier. By default the chat rooms were persistent, muc = new MultiUserChat(connection, groupid+"@conference

pyxmpp: quick tutorial for creating a muc client?

删除回忆录丶 提交于 2019-12-03 14:15:50
I'm attempting to write a quick load-test script for our ejabberd cluster that simply logs into a chat room, posts a couple of random messages, then exits. We had attempted this particular test with tsung, but according to the authors, the muc functionality did not make it into this release. pyxmpp seems to have this functionality, but darned if I can figure out how to make it work. Here's hoping someone has a quick explanation of how to build the client and join/post to the muc. Thanks! Richy Hey I stumbled over your question a few times, while trying the same thing. Here is my answer: Using

XMPPFramework - How to Create a MultiUserChat Rooms?

倾然丶 夕夏残阳落幕 提交于 2019-12-01 12:29:45
How can I Achieve GroupChat in iPhone using XMPPFramework.I tried fallowing code but room is not created.How can I know whether room is created or not.XMPPRoomDelegate are not called.When Stream disconnected, handleDidLeaveRoom method called.Can any one help me.thanks inadvance #define XMPP_HOSTNAME_2 @"chat.someservername.com" #define XMPP_JID @"venkat@chat.someservername.com" #define XMPP_PASSWORD @"venkat" #define ROOM_JID @"venkat_muc@conference.chat.someservername.com/iMac" - (void)mucSetupStream { xmppStream = [[XMPPStream alloc] init]; xmppStream.hostName = XMPP_HOSTNAME_2; xmppStream

Accepting chatroom invitation

孤人 提交于 2019-11-28 18:59:16
I'm able to create a MUC using XMPPFramework and send user invitation requests to join that room by using the code below. // Creating AppDelegate *dele =(AppDelegate *) [[UIApplication sharedApplication]delegate]; xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:dele jid:[XMPPJID jidWithString:self.roomName] dispatchQueue:dispatch_get_main_queue()]; [xmppRoom addDelegate:dele delegateQueue:dispatch_get_main_queue()]; [xmppRoom activate:dele.xmppStream]; [xmppRoom joinRoomUsingNickname:self.myNick history:nil]; // Inviting [xmppRoom inviteUser:[XMPPJID jidWithString:@"abc@host"] withMessage:@