[undertow-dev] Correctly shutting down a websocket handler

[Stuart Douglas]

@OnError should be called if @OnClose is not called (if neither is being called this is a bug). 

Stuart

----- Original Message -----
> From: "Dennis Gesker" <dennis at gesker.com>
> To: "Robin Anil" <robin.anil at gmail.com>
> Cc: undertow-dev at lists.jboss.org
> Sent: Saturday, 21 November, 2015 3:51:31 AM
> Subject: Re: [undertow-dev] Correctly shutting down a websocket handler
> 
> Hi Robin:
> 
> I'm new to WebSockets but I'm glad to share what has been working for me. My
> use cases are currently via Java SE not the Web so this may not apply. I'm
> migrating a bunch of data from an old system and really just pounding on
> Undertow to see how it will behave when I'm ready to build my production
> application. And, I in no way claim this is a best practice. But, maybe you
> will find it useful.
> 
> Anyway, I add three additional methods to each of my deocorated websockets
> (clients) two of which I call from other objects as needed. With this
> approach I'm not seeing any data loss. Seem to be getting timely responses
> from the server side WebSocket. And, always get a nice clean [1000] close
> codes.
> 
> setup()
> tearDown()
> sendWorktoServer()
> 
> 
> public void sendWorkToServer(Object obj) {
> // logger.info (this.getClass().getSimpleName() + " >>> Enter
> sendWorkToServer");
> 
> if (webSocketSession == null || !webSocketSession.isOpen()) {
> setUp();
> }
> 
> if (obj == null || obj.getId() == null || obj.getId().isEmpty()) {
> return;
> }
> 
> ObjectMapper mapper = new ObjectMapper();
> String json = null;
> 
> 
> 
> try {
> json = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(obj);
> logger.info (json);
> webSocketSession.getAsyncRemote().sendText(json);
> //messageLatch.await(500, TimeUnit.MILLISECONDS);
> } catch (JsonProcessingException e) {
> e.printStackTrace();
> }
> 
> 
> logger.info ("Socket is open: " + webSocketSession.isOpen());
> logger.info ("Exit sendWorkToServer");
> }
> 
> private boolean setUp() { //Called if needed inside sendWorkToServer() method
> 
> try {
> websocketServerURI = serverSocketURI.getAddCfgStructureWsUri(); // Get
> Correct URI Development vs Production
> logger.info ("URI: " + websocketServerURI.toString());
> webSocketContainer = ContainerProvider.getWebSocketContainer(); // WebSocket
> Container
> webSocketSession = webSocketContainer.connectToServer(this.getClass(),
> websocketServerURI); // Connect to the Server
> if (!webSocketSession.isOpen()) {
> messageLatch.await(10, TimeUnit.SECONDS);
> }
> 
> } catch (DeploymentException e) {
> e.printStackTrace();
> return false;
> } catch (IOException e) {
> e.printStackTrace();
> return false;
> } catch (InterruptedException e) {
> e.printStackTrace();
> }
> 
> return true;
> 
> }
> 
> public boolean tearDown() {
> 
> if (webSocketSession != null) {
> if (webSocketSession.isOpen()) {
> try {
> webSocketSession.close();
> } catch (IOException e) {
> e.printStackTrace();
> return false;
> }
> }
> }
> webSocketSession = null;
> webSocketContainer = null;
> websocketServerURI = null;
> serverSocketURI = null;
> return true;
> }
> 
> 
> So, inside my other object the code looks something like...
> 
> WebSocketClient wcs = new WebSocketClient();
> wcs.sendWorkToServer(Object obj);
> // More work and activities
> wcs.teardown();
> wcs = null;
> 
> Also, what might be helpful is to add a custom method that just return the
> status of your websocketclient session and then call teardown() as needed.
> I hope that helps a bit... In my scenario I take some extra steps to control
> the life cycle so I don't really have to detect status.
> 
> Cordially,
> Dennis
> 
> 
> 
> 
> 
> 
> 
> On Thu, Nov 19, 2015 at 1:12 PM, Robin Anil < robin.anil at gmail.com > wrote:
> 
> 
> 
> When a client disconnects, I see that onClose is not being fired. The only
> way this seems to be firing if client sents a close frame.
> 
> Is there any way to detect disconnection and immediately close all the opened
> resources.
> 
> Robin
> 
> Robin Anil | Software Engineer
> 
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> 
> 
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> 
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> may love thee. Speak the truth always, even if it leads to your death.
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