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zycoder #1
Member since Dec 2017 · 4 posts
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Subject: how to create a new durable room?
Now,i want to create a new durable room,and codes are

  1. private static XmppClient xmppClient = new XmppClient { XmppDomain = "172.16.100.121", Username = "zhangy", Password = "admin" };
  2.         static void Main(string[] args)
  3.         {
  4.  
  5.             xmppClient.Open();
  6.             MucManager muc = new MucManager(xmppClient);
  7.             Jid jid = new Jid("test@conference.172.16.100.121");
  8.             muc.EnterRoom("xiao@conference.172.16.100.121", "HostUser", true);
  9.  
  10.  
  11.         }

but they can't to work
so can you help me?
Thanks!
This post was edited 2 times, last on 2017-12-08, 12:16 by Alex.
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Alex #2
Member since Feb 2003 · 4327 posts · Location: Germany
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take a look at the MucManager class.
here in this blog posts are also samples included:

  • https://www.ag-software.net/2013/08/27/await-matrix/
  • https://www.ag-software.net/2014/01/21/task-based-asynchronous-muc-room-configuration/
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zycoder #3
Member since Dec 2017 · 4 posts
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thanks your reply  but now i can't find this methond with name is   
 var nick = await mucManager.DiscoverReservedNicknameAsync(roomJid);
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zycoder #4
Member since Dec 2017 · 4 posts
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In reply to post #2
now, i use muc.EnterRoom("", "");  that i can create a room 
but other can't enter this room?
i want to kown how to do?
and you can help me?
thanks!
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zycoder #5
Member since Dec 2017 · 4 posts
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In reply to post #2
now i have solved my issues
I'm sorry to disturb you again
i what to get all members in a room
What should I do?
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Alex #6
Member since Feb 2003 · 4327 posts · Location: Germany
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Please look at the functionality of the MucManager class.
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