每个 Spring websocket 事件的 Principal 为 null

Question

我正在尝试从 Spring websocket SessionConnectEvent 获取 Principal 用户名，但它在每个侦听器上都为 null。我可能做错了什么？

为了实现它，我遵循了您将在此处找到的答案：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(); // headers.getUser() is null
        log.debug("username is " + username);
    }

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

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

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

}

这是我的安全配置：

@Configuration
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter {

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http.authorizeRequests()
            .anyRequest()
            .permitAll()
            .and().csrf().disable();
    }
}

Answer 1

由于我没有实现身份验证机制，因此 Spring 没有足够的信息来提供 Principal 用户名。所以我要做的就是配置一个 HandshakeHandler 来生成 Principal。

@Configuration
@EnableWebSocketMessageBroker
public class WebSocketConfig implements WebSocketMessageBrokerConfigurer {

    public static final String ENDPOINT_CONNECT = "/connect";
    public static final String SUBSCRIBE_USER_PREFIX = "/private";
    public static final String SUBSCRIBE_USER_REPLY = "/reply";
    public static final String SUBSCRIBE_QUEUE = "/queue";

    @Override
    public void configureMessageBroker(MessageBrokerRegistry registry) {
        registry.enableSimpleBroker(SUBSCRIBE_QUEUE, SUBSCRIBE_USER_REPLY);
        registry.setUserDestinationPrefix(SUBSCRIBE_USER_PREFIX);
    }

    @Override
    public void registerStompEndpoints(StompEndpointRegistry registry) {
        registry.addEndpoint(ENDPOINT_CONNECT)
                // assign a random username as principal for each websocket client
                // this is needed to be able to communicate with a specific client
                .setHandshakeHandler(new AssignPrincipalHandshakeHandler())
                .setAllowedOrigins("*");
    }

}

/**
 * Assign a random username as principal for each websocket client. This is
 * needed to be able to communicate with a specific client.
 */
public class AssignPrincipalHandshakeHandler extends DefaultHandshakeHandler {
    private static final String ATTR_PRINCIPAL = "__principal__";

    @Override
    protected Principal determineUser(ServerHttpRequest request, WebSocketHandler wsHandler, Map<String, Object> attributes) {
        final String name;
        if (!attributes.containsKey(ATTR_PRINCIPAL)) {
            name = generateRandomUsername();
            attributes.put(ATTR_PRINCIPAL, name);
        } else {
            name = (String) attributes.get(ATTR_PRINCIPAL);
        }
        return new Principal() {
            @Override
            public String getName() {
                return name;
            }
        };
    }

    private String generateRandomUsername() {
        RandomStringGenerator randomStringGenerator = 
                new RandomStringGenerator.Builder()
                    .withinRange('0', 'z')
                    .filteredBy(CharacterPredicates.LETTERS, CharacterPredicates.DIGITS).build();
        return randomStringGenerator.generate(32);
    }
}

Answer 2

查看AbstractSubProtocolEvent（您感兴趣的所有事件中的superclass）的实现，您可以看到用户被保存在一个单独的字段中。因此，您可以通过调用event.getUser() 来访问用户。您不需要从消息中获取它。

例如对于SessionConnectedEvent，您可以看到用户在事件中被填充，但不是消息。

更新：

您只有在对 http 升级进行身份验证后才能访问该用户。所以你需要有一个 WebSecurityConfigurerAdapter 来配置类似的东西：

@Configuration
public static class UserWebSecurity extends WebSecurityConfigurerAdapter {

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http.requestMatchers()
                .antMatchers(WebsocketPaths.WEBSOCKET_HANDSHAKE_PREFIX); //You configured the path in WebSocketMessageBrokerConfigurer#registerStompEndpoints

        http
                .authorizeRequests()
                .anyRequest().authenticated();
    }

}