muc | 易学教程

ejabberd MAM does not work for MUC

阅读更多关于 ejabberd MAM does not work for MUC

问题 xep-0313 works for rosters, but does not work for room chat. I can get archive message for the rosters, but I can not get archive message for the room, why? <iq type='set' id='juliet1'> <query xmlns='urn:xmpp:mam:0'> <x xmlns='jabber:x:data' type='submit'> <field var='FORM_TYPE' type='hidden'> <value>urn:xmpp:mam:0</value> </field> <field var='with'> <value>roomJid@conference.host</value> </field> </x> </query> </iq> The receive message is: <message xmlns="jabber:client" from="test3@192.168.1

XMPP Groupchat Bookmark implementation

阅读更多关于 XMPP Groupchat Bookmark implementation

问题 I have read in XMPP that in case of XMPP rooms, once a user bookmarks the room , he will automatically rejoin the room on re login. But in my case bookmark is getting stored in server as when I query it back Im able to retrieve it. But rejoining is not happening. Also for retrieving participants list in a members only private group, members are not able to retrieve admin informations. Is it possible to retrive admin info? 回答1: But in my case bookmark is getting stored in server as when I

Smack - When to call MultiUserChat.Join(…)

阅读更多关于 Smack - When to call MultiUserChat.Join(…)

问题 I'm a bit confused and can't find the answer in the docs. I'm developing a simple chat app for Android. I've created a group (muc) and everything works, but do I have to keep calling join() everytime I open the app? I've noticed that it doesn't work otherwise.. Don't understand why since the server knows that my jabberID is connected to this muc. Hope my question is clear enough. Thanks. 回答1: A (I Hope) complete explanation: short answer is yes. When a persistent room is created on your

Xmpp IOS multiuser chat . i didnot find a way to accept the invitation from group ? how i can accept the incomming invitation

阅读更多关于 Xmpp IOS multiuser chat . i didnot find a way to accept the invitation from group ? how i can accept the incomming invitation

问题 When i send invitation this function is called but i can't understand what line of code should use for accept invitation*. am trying to create a multi user and multi groups invitation also called did received message function. - (void)xmppMUC:(XMPPMUC *) sender roomJID:(XMPPJID *) roomJID didReceiveInvitation:(XMPPMessage *)message { } 回答1: This is how you can accept the group invitation. You just need to activate your XMPPMUC protocol as below: XMPPMUC * xmppMUC = [[XMPPMUC alloc]

Join an existing XMPP MUC Room with Owner Privileges using iOS Openfire and Robbie Hanson library

阅读更多关于 Join an existing XMPP MUC Room with Owner Privileges using iOS Openfire and Robbie Hanson library

问题 I am joining an existing MUC room using the next code: XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init]; XMPPJID *roomJID = [XMPPJID jidWithString:@"newRoom5@conference.administrator"]; xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomStorage jid:roomJID dispatchQueue:dispatch_get_main_queue()]; [xmppRoom activate:[self appDelegate].xmppStream]; [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()]; [xmppRoom joinRoomUsingNickname:[self appDelegate]

Join an existing XMPP MUC Room with Owner Privileges using iOS Openfire and Robbie Hanson library

阅读更多关于 Join an existing XMPP MUC Room with Owner Privileges using iOS Openfire and Robbie Hanson library

I am joining an existing MUC room using the next code: XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init]; XMPPJID *roomJID = [XMPPJID jidWithString:@"newRoom5@conference.administrator"]; xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:roomStorage jid:roomJID dispatchQueue:dispatch_get_main_queue()]; [xmppRoom activate:[self appDelegate].xmppStream]; [xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()]; [xmppRoom joinRoomUsingNickname:[self appDelegate].xmppStream.myJID.user history:nil password:nil]; XMPPMUC * xmppMUC = [[XMPPMUC alloc]

Persistent XMPP MUC (XEP-45), like WhatsApp groupchats

阅读更多关于 Persistent XMPP MUC (XEP-45), like WhatsApp groupchats

问题 From the spec — 7.14 Exiting a Room In order to exit a multi-user chat room, an occupant sends a presence stanza of type "unavailable" to the <room@service/nick> it is currently using in the room. Example 80. Occupant Exits a Room <presence from='hag66@shakespeare.lit/pda' to='coven@chat.shakespeare.lit/thirdwitch' type='unavailable'/> This implies that as soon as the user disconnects from the XMPP server, he is removed from the group on the server side. The issue is simple — I don't want