Scala - 如何使用Await.result在Futures上使用Timer而不阻塞

Question

I have an Rest API provided by akka-http . In some cases I need to get data from an external database ( Apache HBase ), and I would like the query to fail if the database takes too long to deliver the data.

One naïve way is to wrap the call inside a Future and then block it with an Await.result with the needed duration.

import scala.concurrent.duration._
import scala.concurrent.{Await, Future}

object AsyncTest1 extends App {

  val future = Future {
    getMyDataFromDB()
  }

  val myData = Await.result(future, 100.millis)
}

The seems to be inefficient as this implementation needs two threads. Is There an efficient way to do this ?

I have another use case where I want to send multiple queries in parallel and then aggregates the results, with the same delay limitation.

val future1 = Future {
    getMyDataFromDB1()
  }

  val future2 = Future {
    getMyDataFromDB2()
  }

  val foldedFuture = Future.fold(
    Seq(future1, future2))(MyAggregatedData)(myAggregateFunction)
  )

  val myData = Await.result(foldedFuture, 100.millis)

Same question here, what is the most efficient way to implement this ?

Thanks for your help

Answer 1

One solution would be to use Akka's after function which will let you pass a duration, after which the future throws an exception or whatever you want.

Take a look here . It demonstrates how to implement this.

EDIT: I guess I'll post the code here in case the link gets broken in future:

import scala.concurrent._
import scala.concurrent.duration._
import ExecutionContext.Implicits.global
import scala.util.{Failure, Success}
import akka.actor.ActorSystem
import akka.pattern.after

val system = ActorSystem("theSystem")

lazy val f = future { Thread.sleep(2000); true }
lazy val t = after(duration = 1 second, using = system.scheduler)(Future.failed(new TimeoutException("Future timed out!")))

val fWithTimeout = Future firstCompletedOf Seq(f, t)

fWithTimeout.onComplete {
   case Success(x) => println(x)
   case Failure(error) => println(error)
}