我尝试用路由器类和处理器类来制作Spring webflux安全应用程序。首先,下面的代码是webflux安全的配置代码。
@Configuration
@EnableWebFluxSecurity
public class BlogWebFluxSecurityConfig {
@Bean
public MapReactiveUserDetailsService userDetailsService() {
UserDetails userWebFlux = User.withUsername("joseph").password("password").roles("USER").build();
return new MapReactiveUserDetailsService(userWebFlux);
}
@Bean
public SecurityWebFilterChain springSecurityFilterChain(ServerHttpSecurity http) {
http
.authorizeExchange()
.pathMatchers("/route/user/all", "/route/post/all").permitAll()
.pathMatchers(HttpMethod.GET, "/route/user/**", "/route/post/**").hasRole("USER")
.anyExchange().authenticated()
.and()
.httpBasic();
return http.build();
}
}
接下来的代码是关于路由器类的。
@Configuration
@EnableWebFlux
public class BlogWebFluxEndpointRouter {
@Bean
public RouterFunction<ServerResponse> routesUser(UserHandler handler) {
return RouterFunctions.route(RequestPredicates.GET("/route/user/all"), handler::findAll)
.andRoute(RequestPredicates.GET("/route/user/id/{id}"), handler::findById)
.andRoute(RequestPredicates.GET("/route/user/username/{username}"), handler::findByUsername)
.andRoute(RequestPredicates.GET("/route/user/email/{email}"), handler::findByEmail)
.andRoute(RequestPredicates.POST("/route/user/create"), handler::register)
.andRoute(RequestPredicates.GET("/route/user/login/{username}/{password}"), handler::authenticate);
}
@Bean
public RouterFunction<ServerResponse> routesPost(PostHandler handler) {
return RouterFunctions.route(RequestPredicates.GET("/route/post/all"), handler::findAll)
.andRoute(RequestPredicates.GET("/route/post/id/{id}"), handler::findById)
.andRoute(RequestPredicates.GET("/route/post/delete/{id}"), handler::deleteById)
.andRoute(RequestPredicates.POST("/route/post/create"), handler::create)
.andRoute(RequestPredicates.PUT("/route/post/{id}/{content}"), handler::edit);
}
}
即使网络是休息的Web服务,但我使用WebFlux的WebClient类。
public void functionOnUserDocument() {
client.get().uri("/route/user/all").accept(MediaType.APPLICATION_JSON).exchange()
.flatMapMany(response -> response.bodyToFlux(User.class))
.subscribe(u -> System.out.println("All Users : " + u.getUsername() + ":" + u.getEmail() + ":" + u.getFullname()));
client.get().uri("/route/user/id/{id}", "0002").accept(MediaType.APPLICATION_JSON).exchange()
.flatMap(response -> response.bodyToMono(User.class))
.subscribe(u -> System.out.println("GET by Id : " + u.getUsername() + ":" + u.getEmail() + ":" + u.getFullname()));
client.get().uri("/route/user/username/{username}", "jina").accept(MediaType.APPLICATION_JSON).exchange()
.flatMap(response -> response.bodyToMono(User.class))
.subscribe(u -> System.out.println("Get by username : " + u.getUsername() + ":" + u.getEmail() + ":" + u.getFullname()));
client.get().uri("/route/user/email/{email}", "myson@college.ac.kr").accept(MediaType.APPLICATION_JSON).exchange()
.flatMap(response -> response.bodyToMono(User.class))
.subscribe(u -> System.out.println("Get By Email : " + u.getUsername() + ":" + u.getEmail() + ":" + u.getFullname()));
client.get().uri("/route/user/login/{username}/{password}", "julian", "password").exchange()
.map(ClientResponse::statusCode).subscribe(response -> System.out.println("Login : " + response.getReasonPhrase()));
User user = new User("0005", 4L, "jane", "password", "aaa@bbb.com", "누나", "USER");
client.post().uri("/route/user/create").body(Mono.just(user), User.class).exchange()
.map(ClientResponse::statusCode).subscribe(response -> System.out.println("User Creation: " + response.getReasonPhrase()));
}
因为我做了webflux安全配置,肯定有些webclient不能执行,禁止如下,
Login : Unauthorized
User Creation: Forbidden
我不使用curl。所以我想知道的是我的WebClient方法是什么,用户名和密码必须在哪里找到并传递给WebClient类。任何回复都将是感谢的。
3条答案
按热度按时间iecba09b1#
从Spring 5.1开始,您应该使用
HttpHeaders#setBasicAuth
设置基本身份验证,如下所示:以前使用
.filter(basicAuthentication("user", "password")
的方法现在已被弃用。nwlqm0z12#
HTTP基本身份验证要求在
Authorization
标头中使用Base64格式编码的用户名和密码。此外,您不需要具有登录端点,因为此信息应随每个请求一起发送。将Basic Auth标头添加到客户端中的每个调用,如下所示:
dl5txlt93#
Spring提供了API,用于通过ClientFilters向WebClient提供基本的身份验证参数。
您可以使用较少的自定义编码来实现与设置
Authorization
头文件相同的结果。请参阅以下来自Spring文档的代码片段: