java Spring Websockets @SendToUser 无需登录？
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Spring Websockets @SendToUser without login?
提问by kentobi
I have a simple spring application with websocket functionality and everything works so far. Now I want to send a message from my server to a specific client using the @SendToUser annotation. This gives me the error "Ignoring message, no principal info available". I understand that i have no login whatsoever on my server, so every user is "anonymous" and does not have a principal (I am not using spring security for now). But every user has a session-id. Isnt it possible to use the session id somehow to differentiate between users? How can i achieve that so my users get a principal which corresponds to the session-id?
我有一个带有 websocket 功能的简单 spring 应用程序，到目前为止一切正常。现在我想使用 @SendToUser 注释从我的服务器向特定客户端发送一条消息。这给了我错误“忽略消息，没有可用的主要信息”。我知道我的服务器上没有任何登录信息，所以每个用户都是“匿名的”并且没有主体（我现在不使用 spring 安全性）。但是每个用户都有一个会话 ID。是否可以以某种方式使用会话 ID 来区分用户？我怎样才能做到这一点，以便我的用户获得与会话 ID 相对应的主体？
采纳答案by Biju Kunjummen
I think a solution might be to avoid using @SendToUserand use raw SimpMessagingTemplateand to send messages to a destination that you control for open sessions.
我认为一个解决方案可能是避免使用@SendToUser和使用原始数据SimpMessagingTemplate，并将消息发送到您控制的打开会话的目的地。
For eg. assuming that you had some identity for a new websocket session, you can subscribe to a queue with that identifier in the queue name:
例如。假设你有一个新的 websocket 会话的一些身份，你可以订阅一个队列名称中具有该标识符的队列：
stomp.subscribe("/queue/chats" + "-" + mycustomidentifier, onmessage);
Now, on the Spring websocket listener side, you can direct your responses using SimpMessagingTemplate:
现在，在 Spring websocket 侦听器端，您可以使用SimpMessagingTemplate以下命令引导您的响应：
@Controller
public class MyController {
@Autowired
private SimpMessagingTemplate simpMessagingTemplate;
@MessageMapping("/chats")
public void handleChat(@Payload ChatMessage message) {
this.simpMessagingTemplate.convertAndSend("/queue/chats-" + "mycustomidentifier", "[" + getTimestamp() + "]:" + message.getMessage());
}
....
回答by R.S
Use @SendToUserand add "/user/" in front of queue when subscribing (only subscriber side). Rest works magic :-)
@SendToUser订阅时使用并在队列前添加“/user/”（仅限订阅者端）。休息有魔法:-)
Instead of
代替
Java Server: @SendTo("/topic/showResult")
and
和
JS Client: stompClient.subscribe('/topic/showResult', function(calResult){ ....
use:
利用：
Java Server: @SentToUser("/topic/showResult")
Java Server: @SentToUser("/topic/showResult")
and
和
JS Client: stompClient.subscribe('/user/topic/showResult', function(calResult){ ....
JS Client: stompClient.subscribe('/user/topic/showResult', function(calResult){ ....
回答by AndrewL
Building on Biju's answerand using the Stomp generated session id(thanks, mariusz2108 in his answer to a similar question), here's what worked for me (based on the canonical example from Spring)
基于Biju 的回答并使用Stomp 生成的会话 ID（感谢mariusz2108 在他对类似问题的回答中），这对我有用（基于Spring 的规范示例）
SpringFramework client:
SpringFramework 客户端：
private SimpMessagingTemplate template;
@Autowired
public GreetingController(SimpMessagingTemplate template) {
this.template = template;
}
@MessageMapping("/hello")
public void greeting(HelloMessage message, @Header("simpSessionId") String sessionId) throws Exception {
template.convertAndSend("/queue/greeting-"+sessionId, new Greeting("Hello, " + message.getName()));
}
JavaScript client:
JavaScript 客户端：
function connect() {
var socket = new SockJS('/gs-guide-websocket');
stompClient = Stomp.over(socket);
stompClient.connect({}, function (frame) {
var sessionId = /\/([^\/]+)\/websocket/.exec(socket._transport.url)[1];
console.log("connected, session id: " + sessionId);
stompClient.subscribe('/queue/greeting-'+sessionId, function (greeting) {
showGreeting(JSON.parse(greeting.body).content);
});
});
}
Instead of the Stomp session id you could use your web container's Session ID (e.g. JSESSIONID) but now that cookie is not by default accessible from JavaScript(for Tomcat) this is a more difficult prospect.
您可以使用 Web 容器的会话 ID（例如 JSESSIONID）代替 Stomp 会话 ID，但现在默认情况下无法从 JavaScript（对于 Tomcat）访问cookie，这是一个更困难的前景。
回答by airush
Try this. It worked for me
试试这个。它对我有用
@Autowired
private SimpMessagingTemplate messagingTemplate;
@MessageMapping("/getHello")
public void sendReply( MessageHeaders messageHeaders, @Payload String message, @Header(name = "simpSessionId") String sessionId){
messagingTemplate.convertAndSendToUser(sessionId, "/queue/hello", "Hello "+ message, messageHeaders);
}
