Question

我正在学习使用Spring WebSockets，但我无法获得一个非常基本的示例，我收到404错误。我使用http://docs.spring.io/spring/docs/current/spring-framework-reference/html/websocket.html作为参考。

WebSocketConfig：

@Configuration
@EnableWebSocket
public class WebSocketConfig implements WebSocketConfigurer {
    private Logger logger = Logger.getLogger(WebSocketConfig.class);

    @Override
    public void registerWebSocketHandlers(WebSocketHandlerRegistry registry) {
        logger.debug("Registering websocket handler");
        registry.addHandler(gameWebSocketHandler(), "/StreamGame/ws/play");
    }

    @Bean
    public WebSocketHandler gameWebSocketHandler() {
        return new GameWebSocketHandler();
    }
}

WebSocketHandler：

public class GameWebSocketHandler extends TextWebSocketHandler {
    final static Logger logger = Logger.getLogger(GameWebSocketHandler.class);

    public GameWebSocketHandler() {
        logger.debug("Constructor handler");
    }

    @Override
    public void handleTransportError(WebSocketSession session, Throwable exception) throws Exception {
        super.handleTransportError(session, exception);
        logger.error("Transport error: " + exception.getMessage());
    }

    @Override
    public void handleTextMessage(WebSocketSession session, TextMessage message) throws Exception {
        logger.info("Handle message received: " + message.getPayload());
        //session.sendMessage(new TextMessage("Here is your answer"));
    }

    @Override
    public void afterConnectionEstablished(WebSocketSession session) throws Exception {
        super.afterConnectionEstablished(session);
        logger.info("Connection established!");
    }
}

除非我犯了错误，否则我应该能够通过websocket从客户端向服务器发送消息，如下所示：

sock = new WebSocket('ws://localhost:8080/StreamGame/ws/play');

我在连接时收到以下错误：

WebSocket握手期间出错：意外的响应代码：404WrappedWebSocket @ VM84：35（匿名函数）@（索引）：11j @jquery-3.1.1.min.js：2k @jquery-3.1.1.min.js： 2 （索引）：26错误未定义

Web.xml包含以下内容：

   <!-- Processes application requests -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/application-context.xml</param-value>
    </context-param>

    <listener>
       <listener-class>
            org.springframework.web.context.ContextLoaderListener
       </listener-class>
    </listener> 
    <servlet>
        <servlet-name>SpringMVCDispatcherServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/application-context.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <!-- Mapping websocket to spring dispatcher servlet -->
    <servlet-mapping>
        <servlet-name>SpringMVCDispatcherServlet</servlet-name>
        <url-pattern>/ws/*</url-pattern>
    </servlet-mapping>

My Spring应用程序上下文包含：

连接websocket时，我在服务器日志中看到以下内容：

>     2016-11-13 TRACE - 39804 - org.springframework.web.servlet.DispatcherServlet -
> http-nio-8080-exec-2 - Bound request context to thread:
> org.apache.catalina.connector.RequestFacade@df5f117
>     2016-11-13 DEBUG - 39804 - org.springframework.web.servlet.DispatcherServlet -
> http-nio-8080-exec-2 - DispatcherServlet with name
> 'SpringMVCDispatcherServlet' processing GET request for
> [/StreamGame/ws/play]
>     2016-11-13 TRACE - 39804 - org.springframework.web.servlet.DispatcherServlet -
> http-nio-8080-exec-2 - Testing handler map
> [org.springframework.web.socket.server.support.WebSocketHandlerMapping@4e92ebf3]
> in DispatcherServlet with name 'SpringMVCDispatcherServlet'
>     2016-11-13 TRACE - 39804 - org.springframework.web.socket.server.support.WebSocketHandlerMapping
> - http-nio-8080-exec-2 - No handler mapping found for [/play]
>     2016-11-13 WARN  - 39804 - org.springframework.web.servlet.PageNotFound - http-nio-8080-exec-2 -
> No mapping found for HTTP request with URI [/StreamGame/ws/play] in
> DispatcherServlet with name 'SpringMVCDispatcherServlet'

Answer 1

您的websocket网址似乎不正确。您已将 / play 注册为处理程序，但请致电 / StreamGame / play

还要记住，ws连接是异步创建的，所以你必须等待连接握手完成才能发送消息，如下所示：

var ws = new WebSocket("ws://localhost:8080/play");
ws.onopen = function() {
    ws.send('AHOIHOI');
}
ws.onerror = function(error) {
    // handle connection error
}

Answer 2

addHandler函数需要映射到＆＃39; / play＆＃39;，因此不包括根上下文（StreamGame）和调度程序映射（/ ws）。更改addHandler解决了这个问题：

registry.addHandler(gameWebSocketHandler(), "/play");

打开时连接的websocket如下：

sock = new WebSocket('ws://localhost:8080/StreamGame/ws/play');

非常基本的Spring websocket返回404

2 个答案: