遍历 Flux 项目并将它们添加到 Mono 对象

Question

我正在开发需要 id 的 api。对于给定的 id，我想从 s3 下载相关数据并将它们放入一个新对象中，我们称之为数据

class Data {
  private List<S3Object> s3Objects; 
  //getter-setter 
} 


   public Mono<ResponseEntity<Data>> getData(@RequestParam List<String> tagIds){
        Data data = new Data();
        Flux<S3Object> s3ObjectFlux = Flux.fromStream(tagIds.stream())
                .parallel()
                .runOn(Schedulers.boundedElastic())
                .flatMap(id -> fetchResources(id))
                .flatMap(idS3Object -> Mono.just(s3Object))
                .ordered((u1, u2) -> u2.hashCode() - u1.hashCode());

        //how do i add it in data object to convert Mono<Data>?       
    }

Answer 1

您需要将其收集到一个列表中，然后将其映射以创建Data 对象，如下所示：

public Mono<ResponseEntity<Data>> getData(@RequestParam List<String> tagIds){
   
   Flux<S3Object> s3ObjectFlux = Flux.fromStream(tagIds.stream())
           .parallel()
           .runOn(Schedulers.boundedElastic())
           .flatMap(id -> fetchResources(id))
           .flatMap(idS3Object -> Mono.just(s3Object))
           .ordered((u1, u2) -> u2.hashCode() - u1.hashCode());

   Mono<Data> data = s3ObjectFlux.collectList()
           .map(s3Objects -> new Data(s3Objects));  
}

创建一个接受 S3 对象列表的构造函数很有帮助：

class Data {
  private List<S3Object> s3Objects; 

  public Data(List<S3Object> s3Objects) {
      this.s3Objects = s3Objects;
  }
  
  //getter-setter 
}