XMPPFramework - How to create a MUC room and invit

Question

I am using Robbiehanson's iOS XMPPFramework. I am trying to create a MUC room and invite a user to the group chat room but it is not working.

I am using the following code:

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[e-mail]/room" nickName:@"room"];
[room createOrJoinRoom];
[room sendInstantRoomConfig];
[room setInvitedUser:@"[e-mail]"];
[room activate:[self xmppStream]];    
[room inviteUser:jid1 withMessage:@"hello please join."];
[room sendMessage:@"HELLO"];

The user [e-mail] should receive the invite message but nothing is happening.

Any help will be appreciated. :)

Answer 1

After exploring various solutions, I've decided to compile and share my implementation here:

1. Create an XMPP Room:

   XMPPRoomMemoryStorage *roomStorage = [[XMPPRoomMemoryStorage alloc] init];
   
   /** 
    * Remember to add 'conference' in your JID like this:
    * e.g. [e-mail]
    */
   
   XMPPJID *roomJID = [XMPPJID jidWithString:@"[e-mail]"];
   XMPPRoom *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];
   

2. Check if room is successfully created in this delegate:

   - (void)xmppRoomDidCreate:(XMPPRoom *)sender
   

3. Check if you've joined the room in this delegate:

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender
   

4. After room is created, fetch room configuration form:

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender {
       [sender fetchConfigurationForm];
   }
   

5. Configure your room

   /**
    * Necessary to prevent this message: 
    * "This room is locked from entry until configuration is confirmed."
    */
   
   - (void)xmppRoom:(XMPPRoom *)sender didFetchConfigurationForm:(NSXMLElement *)configForm 
   {
       NSXMLElement *newConfig = [configForm copy];
       NSArray *fields = [newConfig elementsForName:@"field"];
   
       for (NSXMLElement *field in fields) 
       {
           NSString *var = [field attributeStringValueForName:@"var"];
           // Make Room Persistent
           if ([var isEqualToString:@"muc#roomconfig_persistentroom"]) {
               [field removeChildAtIndex:0];
               [field addChild:[NSXMLElement elementWithName:@"value" stringValue:@"1"]];
           }
       }
   
       [sender configureRoomUsingOptions:newConfig];
   }
   

   References: XEP-0045: Multi-User Chat, Implement Group Chat

6. Invite users

   - (void)xmppRoomDidJoin:(XMPPRoom *)sender 
   {
       /** 
        * You can read from an array containing participants in a for-loop 
        * and send multiple invites in the same way here
        */
   
       [sender inviteUser:[XMPPJID jidWithString:@"keithoys"] withMessage:@"Greetings!"];
   }
   

There, you've created a XMPP multi-user/group chat room, and invited a user. :)

Answer 2

Check the latest XMPPMUCLight & XMPPRoomLight its similar to Whatsapp and other today's trends social app rooms that don't get destroyed or members kicked when offline or no one in room.

Refer this documentation & mod from MongooseIM

Answer 3

I have the feeling that the first thing to do after alloc-init is to attach it to your xmppStream, so it can use xmppStream to send/receive messages.

More exactly:

XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:@"[e-mail]/room" nickName:@"room"];
[room activate:[self xmppStream]];

//other things (create/config/...)