Sending message to specific user on Spring Websocket

Question

How to send websocket message from server to specific user only?

My webapp has spring security setup and uses websocket. I'm encountering tricky problem trying to send message from server to specific user only.

My understanding from reading the manual is from the server we can do

simpMessagingTemplate.convertAndSend("/user/{username}/reply", reply);

And on the client side:

stompClient.subscribe('/user/reply', handler);

But I could never get the subscription callback invoked. I have tried many different path but no luck.

If I send it to /topic/reply it works but all other connected users will receive it too.

To illustrate the problem I've created this small project on github: https://github.com/gerrytan/wsproblem

Steps to reproduce:

1) Clone and build the project (make sure you're using jdk 1.7 and maven 3.1)

$ git clone https://github.com/gerrytan/wsproblem.git
$ cd wsproblem
$ mvn jetty:run

2) Navigate to http://localhost:8080, login using either bob/test or jim/test

3) Click "Request user specific msg". Expected: a message "hello {username}" is displayed next to "Received Message To Me Only" for this user only, Actual: nothing is received

Airspace

Have you been looking at convertAndSendToUser(String user, String destination, T message) ? docs.spring.io/spring/docs/4.0.0.M3/javadoc-api/org/… – Quaver

Yes tried that too but no luck – Airspace

I've been working on private project and this is the method that we use and it is working for us. I think that one potential problem could be that you subscribe to "/user/reply" and you are sending messages to "/user/{username}/reply". I think that you should remove the {username} part and use the convertAndSendToUser(String user, String destination, T message). – Quaver

Thanks but I tried simpMessagingTemplate.convertAndSendToUser(principal.getName(), "/user/reply", reply); and when the message is sent from server it throws this exception java.lang.IllegalArgumentException: Expected destination pattern "/principal/{userId}/**" – Airspace

@ViktorK. is right, and you were quite close to the right solution. Your subscription on client side was correct, you simply had to try: convertAndSendToUser(principal.getName(), "/reply", reply); – Scalene

Answer 1

Oh, client side no need to known about current user, server will do that for you.

On server side, using following way to send message to a user:

simpMessagingTemplate.convertAndSendToUser(username, "/queue/reply", message);

Note: Using queue, not topic, Spring always using queue with sendToUser

On client side

stompClient.subscribe("/user/queue/reply", handler);

Explain

When any websocket connection is open, Spring will assign it a session id (not HttpSession, assign per connection). And when your client subscribe to an channel start with /user/, eg: /user/queue/reply, your server instance will subscribe to a queue named queue/reply-user[session id]

When use send message to user, eg: username is admin You will write simpMessagingTemplate.convertAndSendToUser("admin", "/queue/reply", message);

Spring will determine which session id mapped to user admin. Eg: It found two session wsxedc123 and thnujm456, Spring will translate it to 2 destination queue/reply-userwsxedc123 and queue/reply-userthnujm456, and it send your message with 2 destinations to your message broker.

The message broker receive the messages and provide it back to your server instance that holding session corresponding to each session (WebSocket sessions can be hold by one or more server). Spring will translate the message to destination (eg: user/queue/reply) and session id (eg: wsxedc123). Then, it send the message to corresponding Websocket session

Underpinnings

Can you please explain how you know the username? in your example you said username is 'admin', you receive the username from user? – Kappenne

Username was obtained from HttpSession when initiate a websocket connection – Underpinnings

please could you give more information about how to set the username?. is there anyway i can send the username on suscription message? – Indomitability

@Andres: You can extends DefaultHandshakeHandler and override method determineUser – Underpinnings

@Andres: Send username on subscription is possible but very insecure. Please don't! – Underpinnings

Elaborated on using determineUser & DefaultHandshakeHandler here : #37854227 – Minivet

Your explanation works great. However where is the queue/reply-user[session id] part in official document? – Larose

@ThanhNguyenVan I am following your answer , would you like to look at my question ? thanks . #49748968 – Anne

I agree with @hbrls, according to the code super.convertAndSend(this.destinationPrefix + user + destination, payload, headers, postProcessor);, they do destinationPrefix, then user, THEN destination. Not sure where the "-user" and stuff is being added? Perhaps different version? – Strode

Hello how do we close the connection. I am getting a warning The web application [ROOT] appears to have started a thread named [MessageBroker-4] but has failed to stop it. This is very likely to create a memory leak. The web application [ROOT] appears to have started a thread named [clientInboundChannel-15] but has failed to stop it. This is very likely to create a memory leak. Stack trace of thread: – Americanist

Be aware that you need to change the WebSocketMessageBrokerConfigurer as well! With config.enableSimpleBroker("/topic", "/queue"); in the configureMessageBroker method it works. – Westlund

Answer 2

Ah I found what my problem was. First I didn't register the /user prefix on the simple broker

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

Then I don't need the extra /user prefix when sending:

convertAndSendToUser(principal.getName(), "/reply", reply);

Spring will automatically prepend "/user/" + principal.getName() to the destination, hence it resolves into "/user/bob/reply".

This also means in javascript I had to subscribe to different address per user

stompClient.subscribe('/user/' + userName + '/reply,...) 

Airspace

Mmm... Is there a way to avoid to setup the userName client-side? By changing that value (for example by using another username), you will be able to see messages of others. – Feigin

See my solution: #25647171 – Feigin

how subscribe for to reply to user from methods annotated with @RequestMapping and also @MessageMapping see here How to subscribe using Sping Websocket integration on specific userName(userId) + get notifications from method anotated with @RequestMapping? – Oxalis

Hey my friend can you tell me this? What is this "user" anyway? I am using Spring Security and my users (username) are email addresses. When each user logs in he gets his JWT token on Spring Security. – Impediment

I faced same problems, searchedfor hours before coming to your answer. ONLY your explination helped me. Thank you so much!! – Chittagong

Answer 3

My solution of that based on Thanh Nguyen Van's best explanation, but in addition I have configured MessageBrokerRegistry:

@Configuration
@EnableWebSocketMessageBroker
public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {

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

Liquorish

Answer 4

I created a sample websocket project using STOMP as well. What I am noticing is that

@Configuration
@EnableWebSocketMessageBroker
public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {

@Override
public void configureMessageBroker(MessageBrokerRegistry config) {
    config.enableSimpleBroker("/topic", "/queue");// including /user also works
    config.setApplicationDestinationPrefixes("/app");
}

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

it works whether or not "/user" is included in config.enableSimpleBroker(...

Dupuis

Answer 5

Exactly i did the same and it is working without using user

@Configuration
@EnableWebSocketMessageBroker  
public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {

    @Override
    public void registerStompEndpoints(StompEndpointRegistry registry) {
       registry.addEndpoint("/gs-guide-websocket").withSockJS();
    }

    @Override
    public void configureMessageBroker(MessageBrokerRegistry config) {
        config.enableSimpleBroker("/topic" , "/queue");
        config.setApplicationDestinationPrefixes("/app");
    }
}

Bali

Hey, where do you ever use this /app? In which case? – Impediment

@FranciscoSouza /app is usually used for sending messages from the client to the server. – Westlund

Answer 6

In the following solution, I wrote code snippets for both the client and backend sides. We need to put /user at the start of the socket's topic for client code. Otherwise, the client can not listen to the socket. Dependency

<dependency>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-starter-websocket</artifactId>
</dependency>

WebSocketConfig.java

package com.oktaykcr.notificationservice.config;

import org.springframework.context.annotation.Configuration;
import org.springframework.messaging.simp.config.MessageBrokerRegistry;
import org.springframework.web.socket.config.annotation.EnableWebSocketMessageBroker;
import org.springframework.web.socket.config.annotation.StompEndpointRegistry;
import org.springframework.web.socket.config.annotation.WebSocketMessageBrokerConfigurer;

@Configuration
@EnableWebSocketMessageBroker
public class WebSocketConfig implements WebSocketMessageBrokerConfigurer {

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

    @Override
    public void configureMessageBroker(MessageBrokerRegistry registry) {
        registry.enableSimpleBroker("/file");
        registry.setApplicationDestinationPrefixes("/app");
    }
}

WebSocketContoller.java

package com.oktaykcr.notificationservice.controller;

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.messaging.handler.annotation.MessageMapping;
import org.springframework.messaging.handler.annotation.Payload;
import org.springframework.messaging.handler.annotation.SendTo;
import org.springframework.stereotype.Controller;

@Controller
public class WebSocketController {

    Logger logger = LoggerFactory.getLogger(WebSocketController.class);

    @MessageMapping("/socket")
    @SendTo("/file/status")
    public String fileStatus(@Payload String message) {
        logger.info(message);
        return message;
    }
}

You can send message to socket from anywhere. In my case my principal.getName() is equal to userId.

simpMessagingTemplate.convertAndSendToUser(userId, "/file/status", socketMessage);

Client (ReactJs with react-stomp) App.js

import './App.css';
import SockJsClient from 'react-stomp'
import { useRef } from 'react';

function App() {

  const clientRef = useRef();

  const sendMessage = (msg) => {
    clientRef.current.sendMessage('/app/socket', msg);
  }

  return (
    <div className="App">
      <div>
        <button onClick={() => sendMessage("Hola")}>Send</button>
      </div>
      <SockJsClient url='http://localhost:9090/notification-service/socket' topics={['/user/file/status']}
        onMessage={(msg) => { console.log(msg); }}
        ref={(client) => { clientRef.current = client }} />
    </div>
  );
}

export default App;

The url attribute of the SockJsClient element is http://localhost:9090/notification-service/socket. http://localhost:9090 is Api Gateway ip address, notification-service is name of the microservice and /socket is defined in the WebSocketConfig.java.

Cabalist