我正在开发一个Spring-MVC应用程序，其中我们具有用于身份验证和授权的Spring安全性。我们正在努力迁移到Spring websockets，但是在将经过身份验证的用户放入websocket连接中时遇到了问题。安全上下文在websocket连接中根本不存在，但可以与常规HTTP一起很好地工作。我们在做什么错了？
WebsocketConfig：
@Configuration
@EnableWebSocketMessageBroker
public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {
@Override
public void configureMessageBroker(MessageBrokerRegistry config) {
config.enableSimpleBroker("/topic");
config.setApplicationDestinationPrefixes("/app");
}
@Override
public void registerStompEndpoints(StompEndpointRegistry registry) {
registry.addEndpoint("/app").withSockJS();
}
}
在下面的控制器中，我们试图获取当前经过身份验证的用户，并且该用户始终为空
@Controller
public class OnlineStatusController extends MasterController{
@MessageMapping("/onlinestatus")
public void onlineStatus(String status) {
Person user = this.personService.getCurrentlyAuthenticatedUser();
if(user!=null){
this.chatService.setOnlineStatus(status, user.getId());
}
}
}
security-applicationContext.xml：
<security:http pattern="/resources/**" security="none"/>
<security:http pattern="/org/**" security="none"/>
<security:http pattern="/jquery/**" security="none"/>
<security:http create-session="ifRequired" use-expressions="true" auto-config="false" disable-url-rewriting="true">
<security:form-login login-page="/login" username-parameter="j_username" password-parameter="j_password"
login-processing-url="/j_spring_security_check" default-target-url="/canvaslisting"
always-use-default-target="false" authentication-failure-url="/login?error=auth"/>
<security:remember-me key="_spring_security_remember_me" user-service-ref="userDetailsService"
token-validity-seconds="1209600" data-source-ref="dataSource"/>
<security:logout delete-cookies="JSESSIONID" invalidate-session="true" logout-url="/j_spring_security_logout"/>
<security:csrf disabled="true"/>
<security:intercept-url pattern="/cometd/**" access="permitAll" />
<security:intercept-url pattern="/app/**" access="hasAnyRole('ROLE_ADMIN','ROLE_USER')" />
<!-- <security:intercept-url pattern="/**" requires-channel="https"/>-->
<security:port-mappings>
<security:port-mapping http="80" https="443"/>
</security:port-mappings>
<security:logout logout-url="/logout" logout-success-url="/" success-handler-ref="myLogoutHandler"/>
<security:session-management session-fixation-protection="newSession">
<security:concurrency-control session-registry-ref="sessionReg" max-sessions="5" expired-url="/login"/>
</security:session-management>
</security:http>
答案 0 :(得分：1)
我记得在我从事的项目中遇到了同样的问题。由于我无法使用Spring文档找出解决方案-并且关于Stack Overflow的其他答案对我不起作用-我最终创建了一种解决方法。
技巧本质上是强制应用程序对WebSocket连接请求进行身份验证。为此，您需要一个拦截此类事件的类，然后一旦控制了该类，就可以调用身份验证逻辑。
创建一个实现Spring ChannelInterceptorAdapter的类。在此类内部，您可以注入执行实际身份验证所需的任何bean。我的示例使用基本身份验证：
@Component
public class WebSocketAuthInterceptorAdapter extends ChannelInterceptorAdapter {
@Autowired
private DaoAuthenticationProvider userAuthenticationProvider;
@Override
public Message<?> preSend(final Message<?> message, final MessageChannel channel) throws AuthenticationException {
final StompHeaderAccessor accessor = MessageHeaderAccessor.getAccessor(message, StompHeaderAccessor.class);
StompCommand cmd = accessor.getCommand();
if (StompCommand.CONNECT == cmd || StompCommand.SEND == cmd) {
Authentication authenticatedUser = null;
String authorization = accessor.getFirstNativeHeader("Authorization:);
String credentialsToDecode = authorization.split("\\s")[1];
String credentialsDecoded = StringUtils.newStringUtf8(Base64.decodeBase64(credentialsToDecode));
String[] credentialsDecodedSplit = credentialsDecoded.split(":");
final String username = credentialsDecodedSplit[0];
final String password = credentialsDecodedSplit[1];
authenticatedUser = userAuthenticationProvider.authenticate(new UsernamePasswordAuthenticationToken(username, password));
if (authenticatedUser == null) {
throw new AccessDeniedException();
}
SecurityContextHolder.getContext().setAuthentication(authenticatedUser);
accessor.setUser(authenticatedUser);
}
return message;
}
然后，在您的WebSocketConfig类中，您需要注册拦截器。将上述类添加为bean并注册。完成这些更改后，您的课程将如下所示：
@Configuration
@EnableWebSocketMessageBroker
public class WebSocketConfig extends AbstractWebSocketMessageBrokerConfigurer {
@Autowired
private WebSocketAuthInterceptorAdapter authInterceptorAdapter;
@Override
public void configureMessageBroker(MessageBrokerRegistry config) {
config.enableSimpleBroker("/topic");
config.setApplicationDestinationPrefixes("/app");
}
@Override
public void registerStompEndpoints(StompEndpointRegistry registry) {
registry.addEndpoint("/app").withSockJS();
}
@Override
public void configureClientInboundChannel(ChannelRegistration registration) {
registration.setInterceptors(authInterceptorAdapter);
super.configureClientInboundChannel(registration);
}
}
很显然，身份验证逻辑的细节由您决定。您可以调用JWT服务或任何正在使用的服务。
答案 1 :(得分：1)
如果您使用的是SockJS + Stomp并正确配置了安全性，则应该能够通过常规的用户名/密码验证器（例如@AlgorithmFromHell）进行连接
accessor.setUser(authentication.getPrincipal()) // stomp header accessor
您还可以通过http：// {END_POINT} / access_token = {ACCESS_TOKEN}进行连接。 Spring Security应该能够选择它并通过ResourceServerTokenServices进行loadAuthentication（access_token）。完成此操作后，可以通过将其添加到AbstractSessionWebSocketMessageBrokerConfigurer或WebSocketMessageBrokerConfigurer的印象中来获取主体。这样做时，由于某种原因，加载的Pricipal会保存在“ simpUser”标头中。
@Override
public void configureClientInboundChannel(ChannelRegistration registration) {
registration.interceptors(new ChannelInterceptor() {
@Override
public Message<?> preSend(final Message<?> message, final MessageChannel channel) {
StompHeaderAccessor accessor = MessageHeaderAccessor.getAccessor(message, StompHeaderAccessor.class);
if (accessor != null && StompCommand.CONNECT.equals(accessor.getCommand())) {
if (message.getHeaders().get("simpUser") != null && message.getHeaders().get("simpUser") instanceof OAuth2Authentication) { // or Authentication depending on your impl of security
OAuth2Authentication authentication = (OAuth2Authentication) message.getHeaders().get("simpUser");
accessor.setUser(authentication != null ? (UserDetails) authentication.getPrincipal() : null);
}
}
return message;
}
});
}