小编A1t*_*t0r的帖子

我正在尝试为我根据这些问题开发的 REST API 创建一个过滤器Best Practice for REST token-based authentication with JAX-RS and Jersey。

问题是我调用过滤器的任何方法似乎都不起作用。

这些是我的课程：

安全.java

@NameBinding
@Retention(RetentionPolicy.RUNTIME)
@Target({ElementType.TYPE, ElementType.METHOD})
public @interface Secured { 

}

验证过滤器.java

@Secured
@Provider
@Priority(Priorities.AUTHENTICATION)
public class AuthenticationFilter implements ContainerRequestFilter{

    @Override
    public void filter(ContainerRequestContext requestContext) throws IOException {
        // Get the HTTP Authorization header from the request
        String authorizationHeader = 
            requestContext.getHeaderString(HttpHeaders.AUTHORIZATION);

        // Check if the HTTP Authorization header is present and formatted correctly 
        if (authorizationHeader == null || !authorizationHeader.startsWith("Bearer ")) {
            throw new NotAuthorizedException("Authorization …

我是反应式编程的新手。我看到可以压缩两个单声道以生成结果：

Mono<Info> info = Mono.just(id).map(this::getInfo).subscribeOn(Schedulers.parallel());

Mono<List<Detail>> detail= Mono.just(petitionRequest).map(this.service::getDetails)
                    .subscribeOn(Schedulers.parallel());

Flux<Generated> flux = Flux.zip(detail, info, (p, c) -> {
            Generated o = Generated.builder().info(c).detail(p).build();
            return o;
        });

据我所知，这会并行化这两个调用并生成对象 Generated 当我调用 flux.blockFirst()

如何将另一个单声道合并到现有的两个单声道以生成结果？Flux.zip 只接受两个单声道。

提前致谢。