Question

我正在尝试使用spring 4.0，jsf和glassfish 4.0构建一个简单的教学websocket应用程序我已经创建了一个maven web项目（因为这个应用程序有另一个Web组件（jsf）），并且从这个应用程序我试图设置一些websockets。

@Configuration
@EnableWebSocket
public class WebSocketConfig implements WebSocketConfigurer {

  @Override
  public void registerWebSocketHandlers(WebSocketHandlerRegistry registry) {      
     registry.addHandler(echoHandler(), "/echo");
}
@Bean
  public WebSocketHandler echoHandler() {
  return new EchoHandler();
  }  
}

和

public class EchoHandler extends TextWebSocketHandler {

  @Override
  public void handleTextMessage(WebSocketSession session, TextMessage message) throws Exception {
      session.sendMessage(message);
  }
}

在客户端一个非常简单的连接：

<script>
    /* <![CDATA[ */

    var endPointURL = "ws://localhost:8080/liveasterisk/echo";

    var chatClient = null;

    function connect () {
        chatClient = new WebSocket(endPointURL);
        chatClient.onmessage = function (event) {
           alert(event);
        };
    }

    function disconnect () {
        chatClient.close();
    }

    function sendMessage() {    
            chatClient.send("xxx");    
    }

        connect();

    /* ]]> */    
    </script>

问题是当connect（）方法触发时，我得到404响应。

我想我必须以某种方式训练jsf来响应握手请求。我的所有* .xhtml都映射到jsf servlet。

那么我在这里失踪了什么？

我已经解决了这个问题：

@ServerEndpoint(value = "/keepalive", configurator = SpringConfigurator.class)
public class KeepAliveEndpoint {

private static Logger logger = Logger.getLogger(KeepAliveEndpoint.class);

@Autowired
private KeepAliveService keepAliveService;

  private List<Session> sessions = new ArrayList<Session>();

  @OnOpen
  public void onOpen(Session session) {    
      sessions.add(session);
      System.out.println("onOpen: " + session.getId()+" list size: " + sessions.size());
  }

  @OnClose
  public void onClose(Session session) {
    System.out.println("onClose: " + session.getId());
    sessions.remove(session);
  }

  @OnMessage
  public void handleMessage(Session session, String message) {    

      try{            
          Long userId = Long.parseLong(message);
          keepAliveService.keepAlive(userId);

      }catch(NumberFormatException nfe){
          try {
            session.getBasicRemote().sendText("Cannot perform live update for your status");
        } catch (IOException e) {               
            e.printStackTrace();
        }

      }

  }

}

所以现在我通过jsf暴露了一个套接字，我可以在这个端点注入@Autowired的“服务”。
并使用此js代码：

  <script type="text/javascript">
      var host = "ws://localhost:8080/myApp/keepalive";
      var wSocket = new WebSocket(host);
      var browserSupport = ("WebSocket" in window) ? true : false;

      // called  body onLoad()
      function initializeReception() {

        if (browserSupport) {

          wSocket.onopen = function(){                        
            setInterval(function(){wSocket.send('<h:outputText value="#{managedBean.userDTO.id}" />')}, 300000);              
          };

            // called when a message is received
          wSocket.onmessage = function(event) {        
            alert(event.data);
          };

          // on error handler
          wSocket.onError = function(event) {        
                alert('An error has occured '+event.data+'.');
          };

          // called when socket closes
          wSocket.onclose = function(){
          // websocket is closed.
          //alert("Connection is closed...");
          };
        }
        else {              
          // The browser doesn't support WebSocket
          alert("WebSocket is NOT supported by your Browser!");
        }
      }  

      initializeReception();            
</script>

Answer 1

以上配置适用于Spring MVC的DispatcherServlet。您是否在Web应用程序中配置了一个？根据servlet映射（上面未显示），您很可能需要在URL中添加一个部分以匹配servlet映射。

更长的解释是@EnableWebSocket创建了一个HandlerMapping，它将“/ echo”映射到WebSocketHandler。 HandlerMapping需要驻留在DispatcherServlet的配置中，以便处理HTTP握手。

websockets spring glassfish 4

那么我在这里失踪了什么？

1 个答案: