如何将流的内容放入val？

Question

我有这样的信息流：

def myStream[T: AS: MAT](source: Source[T, NotUsed]): Future[Seq[T]] = {
    return source.runWith(Sink.seq)
}

def myMethod(colorStream: Source[Color, NotUsed]) {
  val allColors = myStream(colorStream).map(_.toList)

  //how can I actually extract the things from allColors
  //so that I can call my method below? myOtherMethod

  if I do println(allColors.map(println _)) I can print the elements fine
}

def myOtherMethod(colors: Seq[Color] = List.empty()) {
 ...
}

Answer 1

allColors是未来。您需要访问将来要包装的内容才能访问colours：Seq [Color]。试试这个：

allColors.onComplete{
  case Success(list) => myOtherMethod(list)
  case Failure(err) => //handle the error
}