I am trying to implement XEP-0055 which is supported by ejabbed as shown here in supported xeps
This is my request:
XMPPIQ *iq = [[XMPPIQ alloc] init];
[iq addAttributeWithName:#"type" stringValue:#"get"];
[iq addAttributeWithName:#"from" stringValue:#"testuser1#company.com"];
[iq addAttributeWithName:#"to" stringValue:#"company.com"];
[iq addAttributeWithName:#"id" stringValue:#"search1"];
XMPPElement *query = [XMPPElement elementWithName:#"query"];
[query setXmlns:#"jabber:iq:search"];
[iq addChild:query];
[self.xmppStream sendElement:iq];
I am getting this response:
<iq xmlns="jabber:client" from="company.com" to="testuser1#company.com/2834146151141475281662718" type="error" id="search1">
<query xmlns="jabber:iq:search"/>
<error code="501" type="cancel">
<feature-not-implemented xmlns="urn:ietf:params:xml:ns:xmpp-stanzas"/>
</error>
</iq>
Why am I receiving feature-not-implemented? Link to official XEP description.
In ejabberd, the search functionality is delegated to a subdomain, by default using the vjud prefix. Try sending the search request to vjud.company.com instead.
Whenever I receive any new messages, these two methods gets called two times.
- (BOOL)xmppStream:(XMPPStream *)sender didReceiveIQ:(XMPPIQ *)iq
- (void)xmppStream:(XMPPStream *)sender didReceiveMessage:(XMPPMessage *)message
I am sending request to server to get my roster list with the following code.
NSXMLElement *query = [NSXMLElement elementWithName:#"query" xmlns:#"jabber:iq:roster"];
XMPPIQ *iq = [XMPPIQ iq];
[iq addAttributeWithName:#"id" stringValue:#"buddyRequest"];
[iq addAttributeWithName:#"to" stringValue:#"54.186.107.171"];
[iq addAttributeWithName:#"type" stringValue:#"get"];
[iq addChild:query];
[self.xmppStream sendElement:iq];
Could somebody point out to me where the problem might lie?
Thank you.
I think you set the delegate to XMPPStream twice.Remove one and check.Then also u got two response , check the xmppstream object value.is both are different , surely you set twice.
I am creating a XMPP chat app on iOS. I am trying to do is search user over XMPP. I checked the opernfire server and it has the search plugin enabled. I gone through XEP:0055 and found the stanza.
See http://xmpp.org/extensions/xep-0055.html for reference on the format of the XML you should send and expect to receive back.
If you are trying to perform a search, here is an example for searching for a user by email address
NSString *bareJID = [self.xmppStream.myJID bare];
NSXMLElement *query = [NSXMLElement elementWithName:#"query"];
[query addAttributeWithName:#"xmlns" stringValue:#"jabber:iq:search"];
NSXMLElement *email = [NSXMLElement elementWithName:#"email" stringValue:searchQuery];
[query addChild:email];
NSXMLElement *iq = [NSXMLElement elementWithName:#"iq"];
[iq addAttributeWithName:#"type" stringValue:#"set"];
[iq addAttributeWithName:#"id" stringValue:#"search2"];
[iq addAttributeWithName:#"to" stringValue:[NSString stringWithFormat:#"search.%#",self.xmppStream.myJID.domain]];
[iq addAttributeWithName:#"from" stringValue:bareJID];
[iq addAttributeWithName:#"xml:lang" stringValue:#"en"];
[iq addChild:query];
[self.xmppStream sendElement:iq];
In your callback for xmppStream:didReceiveIQ: you will receive a response from the server, with the same id you specified above (in this case 'search2'). This will be an XML response, as indicated on the initial URL I mentioned for the XEP-0055 reference page, and you can parse it out accordingly.
You can go http://blog.csdn.net/dangfm/article/details/36185879,and follow '搜索关键词' to assemble XML(iq),then _xmppStream send it(iq).
I am using XMPP in my app. My messages are not being sent whenever I attempt to send a message to any specific ID (for example 'test.codemen#gmail.com' in my code).
My code is given below. Thanks in advance.
- (IBAction)sendMsg:(id)sender
{
NSXMLElement *body = [NSXMLElement elementWithName:#"body"];
[body setStringValue:msgField.text];
NSXMLElement *message = [NSXMLElement elementWithName:#"message"];
[message addAttributeWithName:#"type" stringValue:#"chat"];
[message addAttributeWithName:#"to" stringValue:#"test.codemen#gmail.com"];
[message addChild:body];
iPhoneXMPPAppDelegate *share = [iPhoneXMPPAppDelegate sharedInstance];
[[share xmppStream] sendElement:message];
}
your code looks fine to me.
Google Talk routes messages only to contacts you are subscribed to. Is the user you send the message to on your contact list?
Print the xml , then see what's in the message .Maybe you are sending message to a wrong JID .
I am developing an iOS XMPP chat app that utilizes Robbie Hanson's XMPPFramework.
The most important functionalities have been implemented - sending and receiving messages. Basically, I've built a basic functional chat app already, with a little eye candy of course.
Now, the problem I have is regarding MUC. The codes I saw from other websites show that there is a method initWithRoomName in XMPPRoom. However, this method is absent in the git repo I cloned. So, what is the alternative to this? Or, if there is none, how do I go about creating rooms using XMPPFramework?
Thanks.
Below is how I got my own problem solved. Note that this solution does not involve XMPPRoom at all. First, I created a method that, depending on the situation, either creates or enters a room. (Per XMPP documentation, the XML request for creating is the same is the same as the one you would send to enter a room; that is, if the room has does not exist yet when you enter it, the service will create it for you.)
Here we go. This is the method that creates/enters a room. What this method does is send a presence to the room which you intend to create/enter. If you are the first to enter a room and the room has not been created yet, you automatically become its owner and moderator.
- (void)createOrEnterRoom:(NSString *)roomName
{
//here we enter a room, or if the room does not yet exist, this method creates it
//per XMPP documentation: "If the room does not yet exist, the service SHOULD create the room"
//this method accepts an argument which is what you would baptize the room you wish created
NSXMLElement *presence = [NSXMLElement elementWithName:#"presence"];
NSString *room = [roomName stringByAppendingString:#"#conference.jabber.com/iMac"];
[presence addAttributeWithName:#"to" stringValue:room];
NSXMLElement *x = [NSXMLElement elementWithName:#"x" xmlns:#"http://jabber.org/protocol/muc"];
NSXMLElement *history = [NSXMLElement elementWithName:#"history"];
[history addAttributeWithName:#"maxstanzas" stringValue:#"50"];
[x addChild:history];
[presence addChild:x];
[[self xmppStream] sendElement:presence];
}
Next, in the AppDelegate where XMPPStream methods are declared we filter the XML response we receive in the didReceivePresence method by checking the status code sent by the server. If the status code is 201, bingo! The room creation went just fine. Status codes other than 201 mean different things, but let's focus on 201 for our purpose.
- (void)xmppStream:(XMPPStream *)sender didReceivePresence:(XMPPPresence *)presence
{
NSXMLElement *x = [presence elementForName:#"x" xmlns:#"http://jabber.org/protocol/muc#user"];
for (NSXMLElement *status in [x elementsForName:#"status"])
{
switch ([status attributeIntValueForName:#"code"])
{
case 201: [self notifyRoomCreationOk:room];
}
}
}
Then, we tell the server that what we are creating a room of the type "instant," which means that we will send an IQ element telling it room defaults. notifyRoomCreationOk is a delegate method called in a different view when the room creation succeeds, after all I have to record the room in a text file to make it persistent so that the next time I open the app the room I created before will be visible. In my notifyRoomCreationOk method, I have sendDefaultRoomConfig which, basically, describes what is stated in the first sentence of this paragraph.
-(void)sendDefaultRoomConfig:(NSString *)room
{
NSXMLElement *x = [NSXMLElement elementWithName:#"x" xmlns:#"jabber:x:data"];
[x addAttributeWithName:#"type" stringValue:#"submit"];
NSXMLElement *query = [NSXMLElement elementWithName:#"query" xmlns:#"http://jabber.org/protocol/muc#owner"];
[query addChild:x];
XMPPIQ *iq = [XMPPIQ iq];
[iq addAttributeWithName:#"id" stringValue:[NSString stringWithFormat:#"inroom-cr%#", room]];
[iq addAttributeWithName:#"to" stringValue:room];
[iq addAttributeWithName:#"type" stringValue:#"set"];
[iq addChild:query];
[[self xmppStream ] sendElement:iq];
}
Make sure that you have XMPPStream enabled on the views that call the above methods, otherwise, these won't work. That's all there is to it. Have fun XMPP-ing!
XMPPRoom *room = [[XMPPRoom alloc] initWithRoomName:#"user101#conference.jabber.org/room" nickName:#"room"];
[room createOrJoinRoom];
[room sendInstantRoomConfig];
[room setInvitedUser:#"ABC#jabber.org"];
[room activate:[self xmppStream]];
[room inviteUser:jid1 withMessage:#"hello please join."];
[room sendMessage:#"HELLO"];
the user ABC#jabber.org should receive the invite message
Your post is old, however now I would do it like this:
- (void)createRoomWithJid:(XMPPJID*)roomJID
{
XMPPRoom *xmppRoom = [[XMPPRoom alloc] initWithRoomStorage:self.xmppRoomHybridStorage
jid:roomJID
dispatchQueue:dispatch_get_main_queue()];
[xmppRoom addDelegate:self delegateQueue:dispatch_get_main_queue()];
[xmppRoom activate:self.xmppStream];
[xmppRoom joinRoomUsingNickname:self.xmppStream.myJID.user
history:nil
password:nil];
}
Create chat room by given below code using XMPPFRAMWORK.
let roomStorage: XMPPRoomMemoryStorage = XMPPRoomMemoryStorage()
/**
* Remember to add 'conference' in your JID like this:
* e.g. uniqueRoomJID#conference.yourserverdomain
*/
let roomJID: XMPPJID = XMPPJID.jidWithString("chatRoom_name#conference.myhostname")
let xmppRoom: XMPPRoom = XMPPRoom(roomStorage: roomStorage,
jid: roomJID,
dispatchQueue: dispatch_get_main_queue())
xmppRoom.activate(SKxmpp.manager().xmppStream)
xmppRoom.addDelegate(self, delegateQueue: dispatch_get_main_queue())
xmppRoom.joinRoomUsingNickname(SKxmpp.manager().xmppStream.myJID.user, history: nil, password: nil)
xmppRoom.fetchConfigurationForm()