标签: spring-websocket

我正在将 Spring Web 套接字实现到我们的 Web 应用程序中，并且我想访问套接字控制器方法中的用户名，但我将其设置为 null。

这是代码

@MessageMapping("/user/sockettest" )
@SendTo("/topic/sockettestresult")
public String sockAdd(ListId[] listIds) {
    ..
    SecurityContextHolder.getContext().getAuthentication().getPrincipal();

    return stringRet;
}

spring-security xml 看起来像这样

    <sec:intercept-url pattern="/topic/**" access="permitAll" />

    <sec:intercept-url pattern="/" access="permitAll" />
    <sec:intercept-url pattern="/login*" access="permitAll" />
    <sec:intercept-url pattern="/resources/**" access="permitAll" />

    <sec:form-login login-page="/login" default-target-url="/user/home" always-use-default-target="true" authentication-failure-url="/login?error"/>
    <sec:logout logout-success-url="/login?logout" delete-cookies="JSESSIONID" />
    <sec:access-denied-handler error-page="/denied"/>
    <sec:intercept-url pattern="denied/*" access="permitAll" />

    <sec:csrf disabled="true" />
</sec:http>

套接字配置如下所示

    <websocket:stomp-endpoint path="/user/sockettest">
    <websocket:sockjs/>
    </websocket:stomp-endpoint>

    <websocket:simple-broker prefix="/topic"/>

    <websocket:message-converters register-defaults="false">
        <bean id="mappingJackson2MessageConverter" class="org.springframework.messaging.converter.MappingJackson2MessageConverter">
            <property name="objectMapper" ref="objectMapper"></property>
        </bean>
    </websocket:message-converters>
</websocket:message-broker>

我看过一些示例，其中大多数都在 java 配置中，但是我遇到的问题是，当我使用 …

我有带有网络套接字配置的简单 Spring Boot 应用程序。

当我使用SimpleBroker（由spring提供）运行我的应用程序时，一切正常，但是当我想使用rabbitmq而不是SimpleBroker时，我遇到了一些问题，我的经纪人“不可用”

@Configuration
@EnableWebSocketMessageBroker
open class WebSocketConfig : ILogging by LoggingImp<WebSocketConfig>(), 
WebSocketMessageBrokerConfigurer {

@Autowired
private lateinit var env: Environment

override fun registerStompEndpoints(registry: StompEndpointRegistry) {
    registry.addEndpoint("/hig").setAllowedOrigins("*").withSockJS()
}

override fun configureMessageBroker(registry: MessageBrokerRegistry) {
    registry.setApplicationDestinationPrefixes("/app")
    val host = env.getProperty("spring.rabbitmq.host")!!
    val port = env.getProperty("spring.rabbitmq.port")!!.toInt()
    val login = env.getProperty("spring.rabbitmq.username")!!
    val pass = env.getProperty("spring.rabbitmq.password")!!
    log.debug("webSocket=$host, $port, $login, $pass")
//        registry.enableSimpleBroker("/chat")
    registry.enableStompBrokerRelay("/chat")
            .setRelayHost(host)
            .setRelayPort(port)
            .setClientLogin(login)
            .setClientPasscode(pass)
            .setAutoStartup(true)
            .setSystemHeartbeatReceiveInterval(10000)
            .setSystemHeartbeatSendInterval(10000)
    }

}

来自 WebSocketMessageBrokerStats 的响应：

{
"loggingPeriod": 1800000,
"webSocketSessionStatsInfo": "0 current WS(0)-HttpStream(0)-HttpPoll(0), 0 total, 0 …

在我的 Spring Boot 应用程序中，我也使用了 websockets。一切正常，正如在生产中预期的那样。

现在我开始使用 Spring-Boot-Test 创建单元测试。

每次我启动 @SpringBootTest 时，都会遇到以下异常（缩短）：

java.lang.IllegalStateException: Failed to load ApplicationContext
Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'createWebSocketContainer' defined in class path resource [com/package/spring/ws/MyWebsocketConfig.class]: Invocation of init method failed;
Caused by: java.lang.IllegalStateException: Attribute 'javax.websocket.server.ServerContainer' not found in ServletContext

长版

java.lang.IllegalStateException: Failed to load ApplicationContext

at org.springframework.test.context.cache.DefaultCacheAwareContextLoaderDelegate.loadContext(DefaultCacheAwareContextLoaderDelegate.java:125)
at org.springframework.test.context.support.DefaultTestContext.getApplicationContext(DefaultTestContext.java:108)
at org.springframework.test.context.web.ServletTestExecutionListener.setUpRequestContextIfNecessary(ServletTestExecutionListener.java:190)
at org.springframework.test.context.web.ServletTestExecutionListener.prepareTestInstance(ServletTestExecutionListener.java:132)
at org.springframework.test.context.TestContextManager.prepareTestInstance(TestContextManager.java:246)
at org.springframework.test.context.junit4.SpringJUnit4ClassRunner.createTest(SpringJUnit4ClassRunner.java:227)
at org.springframework.test.context.junit4.SpringJUnit4ClassRunner$1.runReflectiveCall(SpringJUnit4ClassRunner.java:289)
at org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12)
at org.springframework.test.context.junit4.SpringJUnit4ClassRunner.methodBlock(SpringJUnit4ClassRunner.java:291)
at org.springframework.test.context.junit4.SpringJUnit4ClassRunner.runChild(SpringJUnit4ClassRunner.java:246)
at org.springframework.test.context.junit4.SpringJUnit4ClassRunner.runChild(SpringJUnit4ClassRunner.java:97)
at org.junit.runners.ParentRunner$3.run(ParentRunner.java:290)
at org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71)
at org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)
at …

我试图从 Spring websocket SessionConnectEvent 获取主体用户名，但它在每个侦听器上均为空。我可能做错了什么？

为了实现它，我遵循了您可以在这里找到的答案：how to capture connection event in my webSocket server with Spring 4?

@Slf4j
@Service
public class SessionEventListener {

    @EventListener
    private void handleSessionConnect(SessionConnectEvent event) {
        SimpMessageHeaderAccessor headers = SimpMessageHeaderAccessor.wrap(event.getMessage());
        String sessionId = headers.getSessionId();
        log.debug("sessionId is " + sessionId);
        String username = headers.getUser().getName(); // headers.getUser() is null
        log.debug("username is " + username);
    }

    @EventListener
    private void handleSessionConnected(SessionConnectEvent event) {
        SimpMessageHeaderAccessor headers = SimpMessageHeaderAccessor.wrap(event.getMessage());
        String sessionId = headers.getSessionId();
        log.debug("sessionId is " + sessionId);
        String username = headers.getUser().getName(); …

我正在使用 spring WebSocket、SockJs 和 Amazon MQ 构建一个示例聊天应用程序。当客户端订阅主题时，它会抛出“代理不可用”异常。所有入站流量规则都在 AWS 安全组中正确设置，并且代理也有 stomp 支持。我正在关注这个Spring Guide。

如果我使用的是内存代理，它工作正常。我非常感谢您对此的帮助，以下是示例代码。

代理：Amazon MQ（内部使用 Active MQ）

版本：5.15.0

WebSocketConfig.java

@Override
public void configureMessageBroker(MessageBrokerRegistry registry) {

    registry.enableStompBrokerRelay("/topic")
            .setRelayHost("***********.mq.us-east-2.amazonaws.com").setRelayPort(61614)
            .setClientLogin("******").setClientPasscode("*****");

    registry.setApplicationDestinationPrefixes("/app");

}

@Override
public void registerStompEndpoints(StompEndpointRegistry registry) {

    registry.addEndpoint("/chat-endpoint").withSockJS();
}

应用程序启动日志

.......
INFO 14280 --- [alina-utility-1] o.s.m.s.s.StompBrokerRelayMessageHandler : Starting...

INFO 14280 --- [alina-utility-1] o.s.m.s.s.StompBrokerRelayMessageHandler : Starting "system" session, StompBrokerRelay[ReactorNettyTcpClient[reactor.netty.tcp.TcpClientDoOn@7acb7b3e]]

INFO 14280 --- [alina-utility-1] o.s.m.s.s.StompBrokerRelayMessageHandler : Started.
......

客户

var socket = new SockJS('/chat-endpoint');
    stompClient = Stomp.over(socket);

    stompClient.connect({}, function(frame) {

        setConnected(true); …

我正在尝试在我现有的Spring MVC应用程序中实现聊天.我指的是来自github的这两个链接 -

1. 春天STOMP

2. Spring Websocket产品组合

但是当我尝试通过websocket连接时,看起来初始的HttpHandhsake失败了.我已经在这里上传了我的源代码 - 服务器代码.这是关于源代码的简要介绍.配置类是 -

1. ConChactConfiguration
2. ConChactWebSocketConfig

只有ChatController处理WebSocket和webapp/WEB-INF/pages/home.jsp包含客户端js.I我在Sock JS上使用STOMP.

因此,请查看代码并建议我进行聊天工作的更改.目前,当我在网络选项卡上看到它看起来像初始HttpHandshake失败,因为交换机协议请求的延迟非常高(它永远不会返回).

从网络和控制台面板编辑 ScreenShots -

我想创建 2 个网络套接字端点。你能说有可能吗？

在那种情况下应该是什么配置？

如何在WebSocketMessageBrokerStats中设置loggingPeriod以减小值(默认为30')

WebSocketMessageBrkerConfigurationSupport中的@Bean加载WebSocketMessageBrokerStats

版本:Spring 4.2.0.RELEASE

我目前的配置:

@Configuration
@EnableWebSocketMessageBroker
@EnableScheduling
public class AppWebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {

    @Override
    public void configureMessageBroker(final MessageBrokerRegistry config) {
        config.enableSimpleBroker("/topic");
    }

    @Override
    public void registerStompEndpoints(final StompEndpointRegistry registry) {
        registry.addEndpoint("/entry")
                .setAllowedOrigins("*")
                .withSockJS()
                .setDisconnectDelay(10000);
    }

}

当我尝试在Spring websocket上从客户端向服务器发送消息时,我遇到了问题.我在服务器上配置Websocket并在控制器上创建@Message.我通过javascript从客户端发送数据.它有时会工作,但有时会失败并在服务器上抛出消息:MissingSessionUserException:消息中没有"user"标题

这是我的WebsocketConfig:

public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {
    public void registerStompEndpoints(StompEndpointRegistry registry) {
        registry.addEndpoint("/connectsocket").withSockJS();
    }
    public void configureMessageBroker(MessageBrokerRegistry config) {
        config.enableSimpleBroker("/topic");
        config.setApplicationDestinationPrefixes("/");
    }
}

这是我的MessageController:

@RestController
public class MessageController {

    @Autowired
    private SimpMessagingTemplate template;

    @MessageMapping("/websocket/message")
    public synchronized void message(Message<Object> messageObj, 
        WebMessage message, Principal principal) throws Exception {
        if (principal != null) {
            String name = principal.getName();
            template.convertAndSendToUser(name, "/topic/dynamic", new MessagePojo("stage", "value", "message"));
        }
    }
}

这是我的Javascript-backbonejs代码:

app.Models.WebsocketModel = Backbone.Model.extend({
    fetchData : function() {
        console.log("WebsocketModel: fetchData");
        var …

我有一个Spring Websocket Stomp应用程序,它接受SUBSCRIBE请求.

在应用程序中,我有一个SUBSCRIBE的处理程序,即

 @Component
 public class SubscribeStompEventHandler implements ApplicationListener<SessionSubscribeEvent> {

    @Override
    public void onApplicationEvent(SessionSubscribeEvent event) {}
 }

我用来验证订阅.

我会检查onApplicationEvent中的某些内容并从此函数将STOMP ERROR消息发送回客户端.

我找到了这个配方如何使用Spring WebSocket向STOMP客户端发送ERROR消息？但我需要了解如何获得outboundChannel.

我还尝试了以下代码:

   public void sendStompError(SimpMessagingTemplate simpMessagingTemplate, String sessionId, String topic, String errorMessage) {

    StompHeaderAccessor headerAccessor = StompHeaderAccessor.create(StompCommand.ERROR);
    headerAccessor.setMessage(errorMessage);
    headerAccessor.setSessionId(sessionId);
    headerAccessor.setLeaveMutable(true);
    simpMessagingTemplate.convertAndSendToUser(sessionId, topic, new byte[0], headerAccessor.getMessageHeaders());       
}

我试过主题成为一些subsciption主题和/ queue/error主题.但是我没有看到消息传播到客户端.

在客户端,我使用:

      stompClient.connect(headers
                , function (frame) {
                    console.log("Conn OK " + url);               
                }, function (error) {                       
                    console.log("Conn NOT OK " + url + ": " …