如何在 RSocketSecurity(Spring) 中创建路径 public

How to make the path public in RSocketSecurity(Spring)

我有 RSocketSecurity 配置 class 类似的东西

@Configuration
@EnableRSocketSecurity
@EnableReactiveMethodSecurity
class RSocketAuthConfiguration {

及其授权(只允许经过身份验证的用户订阅)

     security.addPayloadInterceptor(interceptor).authorizePayload {
        it.setup().authenticated().anyRequest().permitAll()
    }

我想设置一些具有public访问权限的路由,但其中大部分应该是经过授权的。实现该目标的最佳方法是什么?

以下几行内容应该有效:

@Configuration
@EnableRSocketSecurity
@EnableReactiveMethodSecurity
class RSocketSecurityConfiguration(val authenticationService: AuthenticationService) {

    @Bean
    fun authorization(security: RSocketSecurity): PayloadSocketAcceptorInterceptor {
        return security
                .authorizePayload {
                    it.route("route-A").hasRole("role-A")
                        .route("route-B").permitAll()
                }
                .simpleAuthentication(Customizer.withDefaults())
                .authenticationManager(authenticationService)
                .build()
    }
}

route-A 已通过身份验证并需要 role-Aroute-B 是公开可用的。

Spring Security Rsocket 分别配置 setuproute

下面是配置部分的例子

@Bean
public PayloadSocketAcceptorInterceptor rsocketInterceptor(RSocketSecurity rsocket) {
        return rsocket
                .authorizePayload(
                        authorize -> {
                            authorize
                                    // must have ROLE_SETUP to make connection
                                    .setup().hasRole("SETUP")
                                    // must have ROLE_ADMIN for routes starting with "greet."
                                    .route("greet*").hasRole("ADMIN")
                                    // any other request must be authenticated for
                                    .anyRequest().authenticated();
                        }
                )
                .basicAuthentication(Customizer.withDefaults())
                .build();
    }

从我的 Github 中获取 the complete example