在处理Seq对象时如何链接Futures?

时间:2018-01-15 15:38:07

标签: scala sequence traversal

import scala.concurrent.duration.Duration
import scala.concurrent.duration.Duration._
import scala.concurrent.{Await, Future}
import scala.concurrent.Future._
import scala.concurrent.ExecutionContext.Implicits.global

object TestClass {

  final case class Record(id: String)

  final case class RecordDetail(id: String)

  final case class UploadResult(result: String)

  val ids: Seq[String] = Seq("a", "b", "c", "d")

  def fetch(id: String): Future[Option[Record]] = Future {
    Thread sleep 100
    if (id != "b" && id != "d") {
      Some(Record(id))
    } else None
  }

  def fetchRecordDetail(record: Record): Future[RecordDetail] = Future {
    Thread sleep 100
    RecordDetail(record.id + "_detail")
  }

  def upload(recordDetail: RecordDetail): Future[UploadResult] = Future {
    Thread sleep 100
    UploadResult(recordDetail.id + "_uploaded")
  }

  def notifyUploaded(results: Seq[UploadResult]): Unit = println("notified " + results)

  def main(args: Array[String]): Unit = {

    //for each id from ids, call fetch method and if record exists call fetchRecordDetail 
    //and after that upload RecordDetail, collect all UploadResults into seq
    //and call notifyUploaded with that seq and await result, you should see "notified ...." in console


    // In the following line of code how do I pass result of fetch to fetchRecordDetail function
    val result = Future.traverse(ids)(x => Future(fetch(x)))
    // val result: Future[Unit] = ???

    Await.ready(result, Duration.Inf)
  }

}

我的问题是,我不知道在main中放入哪些代码以使其按照评论中的说明运行。总而言之,我有一个ids:Seq[String],我希望每个id都通过异步方法fetchfetchRecordDetailupload,最后是整个Seq来来notifyUploaded

1 个答案:

答案 0 :(得分:2)

我认为最简单的方法是:

  def main(args: Array[String]): Unit = {

    //for each id from ids, call fetch method and if record exists call fetchRecordDetail
    //and after that upload RecordDetail, collect all UploadResults into seq
    //and call notifyUploaded with that seq and await result, you should see "notified ...." in console

    def runWithOption[A, B](f: A => Future[B], oa: Option[A]): Future[Option[B]] = oa match {
      case Some(a) => f(a).map(b => Some(b))
      case None => Future.successful(None)
    }

    val ids: Seq[String] = Seq("a", "b", "c", "d")

    val resultSeq: Seq[Future[Option[UploadResult]]] = ids.map(id => {
      for (or: Option[Record] <- fetch(id);
           ord: Option[RecordDetail] <- runWithOption(fetchRecordDetail, or);
           our: Option[UploadResult] <- runWithOption(upload, ord)
      ) yield our
    })

    val filteredResult: Future[Seq[UploadResult]] = Future.sequence(resultSeq).map(s => s.collect({ case Some(ur) => ur }))
    val result: Future[Seq[UploadResult]] = filteredResult.andThen({ case Success(s) => notifyUploaded(s) })

    Await.ready(result, Duration.Inf)
  }

这个想法是你首先通过所有方法获得Seq[Future[_]] map(这里使用for-comprehension完成)。这是一个重要的技巧是实际传递Seq[Future[Option[_]]]。通过Option[_]辅助方法将runWithOption传递到整个链中可以大大简化代码,而无需阻止直到最后阶段。

然后,您将Seq[Future[_]]转换为Future[Seq[_]],并过滤掉ids阶段失败的fetch的结果。最后你申请notifyUploaded

P.S。请注意,此代码中没有错误处理,并且不清楚在不同阶段出现错误时如何表现它。