使用 Future.failed 抛出异常或使任务失败答案

【问题标题】：Throwing exception or fail the task using Future.failed使用 Future.failed 抛出异常或使任务失败
【发布时间】：2022-01-19 02:15:53
【问题描述】：

我正在使用 scala 期货来轮询 API 状态。继续轮询直到返回成功或失败。但是，当任何批处理失败时，它应该抛出错误并停止程序。

我无法通过 throw (exception) 或 Future.successful(false) 或 Future.failed(new Exception("error")) 抛出错误。


package FutureScheduler

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


object delayTest extends App {

  import scala.concurrent.Future
  import scala.concurrent.duration._
  import scala.concurrent.ExecutionContext.Implicits.global


  def getStatus(batchId: String): String = {
    print(s" $batchId ${System.currentTimeMillis()} continue \n")
    batchId match {
      case "batch1" => "success"
      case "batch2" => "failed"
      case "batch2" => "success" 
    }
  }


  def waitTask(batch: (String, Int)
              ): Future[Boolean] =
    Delayed(x.seconds)(getStatus(batch._1)).flatMap {
      case "success" =>
        print(s"\n${batch._1} succeeded for ${batch._2}")
        Future.successful(true)
      case "failed"  =>
        print(s"\n${batch._1} failed for ${batch._2}")
        Future.failed(new Exception("error"))
        throw new RuntimeException("errored")
      case _ => {
        waitTask(batch)
      }
    }


  val statusList =  List(Some("batch1", 123), None, Some("batch2", 124)).flatten

  val y = 1
  val x = 5

  try {
    Await.ready(Future.traverse(statusList)((waitTask _)), y.minutes)
  }
  catch {
    case e: Exception => println("caught error")
  }

  print("\nbye now")

}

import scala.concurrent.duration._

import scala.concurrent.{Future, Promise}

object Delayed {
  import java.util.{Timer, TimerTask}
  private val timer = new Timer
  def apply[T](delay: Duration)(task: => T): Future[T] = {
    val promise = Promise[T]()
    val tt = new TimerTask {
      override def run(): Unit = promise.success(task)
    }
    timer.schedule(tt, delay.toMillis)
    promise.future
  }
}

【问题讨论】：

Delayed 定义在哪里？

标签： scala

【解决方案1】：

throw 发生在由Delayed 返回的Future 内，因此它会被Future 捕获。

你需要把Await.ready变成Await.result，然后看它返回的值，得到测试的结果。

【讨论】：