如何使用Spring WebSocket向STOMP客户端发送错误消息?

我正在使用Spring的STOMP over WebSocket实现,以及一个功能齐全的ActiveMQ代理。当用户访问某个主题时,他们必须通过一些权限逻辑才能成功订阅。我正在使用通道接收器来应用权限逻辑,如下配置:SUBSCRIBE

WebSocketConfig.java:

@EnableWebSocketMessageBroker
public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {

  @Override
  public void registerStompEndpoints(StompEndpointRegistry registry) {
    registry.addEndpoint("/stomp")
      .setAllowedOrigins("*")
      .withSockJS();
  }

  @Override
  public void configureMessageBroker(MessageBrokerRegistry registry) {
    registry.enableStompBrokerRelay("/topic", "/queue")
      .setRelayHost("relayhost.mydomain.com")
      .setRelayPort(61613);
  }

  @Override
  public void configureClientInboundChannel(ChannelRegistration registration) {
    registration.setInterceptors(new MySubscriptionInterceptor());
  }


}

WebSocketSecurityConfig.java:

public class WebSocketSecurityConfig extends AbstractSecurityWebSocketMessageBrokerConfigurer {

  @Override
  protected void configureInbound(MessageSecurityMetadataSourceRegistry messages) {
    messages
      .simpSubscribeDestMatchers("/stomp/**").authenticated()
      .simpSubscribeDestMatchers("/user/queue/errors").authenticated()
      .anyMessage().denyAll();
  }

}

MySubscriptionInterceptor.java:

public class MySubscriptionInterceptor extends ChannelInterceptorAdapter {

  @Override
  public Message<?> preSend(Message<?> message, MessageChannel channel) {

    StompHeaderAccessor headerAccessor= StompHeaderAccessor.wrap(message);
    Principal principal = headerAccessor.getUser();

    if (StompCommand.SUBSCRIBE.equals(headerAccessor.getCommand())) {
      checkPermissions(principal);
    }

    return message;
  }

  private void checkPermissions(Principal principal) {
    // apply permissions logic
    // throw Exception permissions not sufficient
  }
}

当没有足够的权限的客户端尝试订阅受限制的主题时,它们实际上从未收到来自该主题的任何消息,但也不会收到有关引发的异常的通知,该异常拒绝了其订阅。相反,客户端会返回一个死订阅,ActiveMQ 代理对此一无所知。(与 STOMP 端点和主题的正常、经过充分许可的客户端交互按预期工作。

我尝试订阅,并在成功连接后使用我的Java测试客户端进行简单操作,但是到目前为止,我无法从交付到客户端的服务器获得有关订阅异常的错误消息。这显然不太理想,因为客户端从未收到过他们被拒绝访问的通知。users/{subscribingUsername}/queue/errorsusers/queue/errors


答案 1

你不能只是从 上抛出异常,因为最后一个是 ,因此是,并且来自这些线程的任何异常最终都会在日志中重新抛出给调用方 - 。MySubscriptionInterceptorclientInboundChannelExecutorSubscribableChannelasyncStompSubProtocolHandler.handleMessageFromClient

但是你可以做的事情是这样的,并像这样使用它:clientOutboundChannel

StompHeaderAccessor headerAccessor = StompHeaderAccessor.create(StompCommand.ERROR);
headerAccessor.setMessage(error.getMessage());

clientOutboundChannel.send(MessageBuilder.createMessage(new byte[0], headerAccessor.getMessageHeaders()));

要考虑的另一个选项是注释映射:

    @SubscribeMapping("/foo")
    public void handleWithError() {
        throw new IllegalArgumentException("Bad input");
    }

    @MessageExceptionHandler
    @SendToUser("/queue/error")
    public String handleException(IllegalArgumentException ex) {
        return "Got error: " + ex.getMessage();
    }

答案 2

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