每个Spring websocket事件的主体都为空
每个Spring websocket事件的主体都为空
提问于 2019-06-07 23:14:35
我试图从Spring websocket SessionConnectEvent获取主体用户名，但在每个监听器上都是空的。我能做错什么吗？
为了实现它，我遵循了你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(); // 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();
}
}回答 2
Stack Overflow用户
回答已采纳
发布于 2019-06-11 01:44:05
由于我没有实现身份验证机制，Spring没有足够的信息来提供一个主体用户名。所以我要做的就是配置一个生成主体的HandshakeHandler。
@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);
}
}Stack Overflow用户
发布于 2019-06-08 03:13:49
查看AbstractSubProtocolEvent (您感兴趣的所有事件的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()来简单地访问用户。你不需要从消息中得到它。
例如，对于SessionConnectedEvent，您可以看到用户在事件中填充，而不是在消息中填充。
更新：
只有在对http升级进行身份验证后，才能访问该用户。因此，您需要有一个配置如下内容的WebSecurityConfigurerAdapter：
@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();
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56497055
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