groupchat

How to send composing / is typing (Chat states) event in Multiuser chat / Group chat & One to One chat in xmpp android?

由 时光总嘲笑我的痴心妄想 提交于 2020-01-01 03:47:05
问题 I want send composing event in Group (Multiuser) chat in xmpp, I am using asmack library, I have done same functionality with One to One chat. I am using below code: mMessageEventManager = new MessageEventManager(XMPPConnectApplication.getInstance().getXmppConnection()); mMessageEventManager.addMessageEventNotificationListener(new MessageEventNotificationListener() { @Override public void offlineNotification(String arg0, String arg1) { } @Override public void displayedNotification(String arg0

Can't able to receive group chat messages using smack-android:4.1.4

由 岁酱吖の 提交于 2019-12-14 02:08:55
问题 I am developing chat app using smack library. I have an issue in group chat. In my app, i am creating a group and in that members are auto-joined.i want to notify all user when I send a message in the group even if they had not initiated a chat.My code is as follow in that I have place listener in init method but unable to receive a message. multiUserChatManager = MultiUserChatManager.getInstanceFor(mConnection); mMultiUserChat = multiUserChatManager.getMultiUserChat(to); mConnection

Getting MucConfigurationNotSupportedException In android smack 4.2.0-beta1

由 ≯℡__Kan透↙ 提交于 2019-12-13 06:31:25
问题 I am developing chat application for one to one chat and group chat. I have successfully done one to one chat. Using the below link I have created Group chat. Link to create Group chat in smack 4.2.0-beta1 I can see the group in admin panel but There is only a single user available, But I have created this group with three members. Here I have added my code. public void createGroupChat() { String DomainName = "conference."+ServiceAddress; // Create a MultiUserChat using a Connection for a

XMPP Groupchat Bookmark implementation

由 冷暖自知 提交于 2019-12-13 05:19:32
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

How to create xmpp group chat and add members using smack 4.1 in android [closed]

由 柔情痞子 提交于 2019-12-08 14:20:41
问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 3 years ago . In my app, I want to create groups and add users to the groups using smack 4.1. I am able to create group but unable to add user in that groups. My Code for create groups is... create group if (connection != null) { String groupName = XmppUtils.buildMUCName(grpName); MultiUserChat multiUserChat =

How can I implement auto join for XMPP room in iOS

由 匆匆过客 提交于 2019-12-07 17:15:53
问题 I want to do Auto Join room in XMPP group chat. I have all of group created from back end Server, first time i join the room but when i disconnect from XMPP (out of the app) and back inside app and re-connect xmpp, I again need to join the room. In side app one user already available in multiple room so not possible to Join again n again in the room . Please help.. Thanks in Advance! 回答1: I think you have to show this link : https://xmpp.org/extensions/xep-0048.html . Here you can find the

How can I implement auto join for XMPP room in iOS

由 扶醉桌前 提交于 2019-12-05 18:52:58
I want to do Auto Join room in XMPP group chat. I have all of group created from back end Server, first time i join the room but when i disconnect from XMPP (out of the app) and back inside app and re-connect xmpp, I again need to join the room. In side app one user already available in multiple room so not possible to Join again n again in the room . Please help.. Thanks in Advance! I think you have to show this link : https://xmpp.org/extensions/xep-0048.html . Here you can find the solution of Auto join and room storage. There is No way for , reEnter to your rooms Automatically. you should

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

由 只谈情不闲聊 提交于 2019-11-27 20:59:08
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

how to get all child list from Firebase android

由 半城伤御伤魂 提交于 2019-11-26 11:16:42
I want all child list from Firebase in android. I have implement this code but its not working. mFirebaseRef = new Firebase(FIREBASE_URL); mFirebaseRef.addValueEventListener(new ValueEventListener() { @Override public void onDataChange(DataSnapshot dataSnapshot) { List<String> td = (ArrayList<String>) dataSnapshot.getValue(); //notifyDataSetChanged(); } @Override public void onCancelled(FirebaseError firebaseError) { } }); I hope below code works Firebase ref = new Firebase(FIREBASE_URL); ref.addValueEventListener(new ValueEventListener() { @Override public void onDataChange(DataSnapshot

how to get all child list from Firebase android

由 我是研究僧i 提交于 2019-11-26 02:19:56
问题 I want all child list from Firebase in android. I have implement this code but its not working. mFirebaseRef = new Firebase(FIREBASE_URL); mFirebaseRef.addValueEventListener(new ValueEventListener() { @Override public void onDataChange(DataSnapshot dataSnapshot) { List<String> td = (ArrayList<String>) dataSnapshot.getValue(); //notifyDataSetChanged(); } @Override public void onCancelled(FirebaseError firebaseError) { } }); 回答1: I hope below code works Firebase ref = new Firebase(FIREBASE_URL)
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