scala - Scala 中压平 'reactive' 调用结果

Question

通过重复链接异步调用，我最终得到了这种可怕的类型:

Seq[Future[Option[Future[T]]]]

有没有办法将其简化为更易于管理:

Future[Seq[T]]

如:

def simplify[T](x: Seq[Future[Option[Future[T]]]]): Future[Seq[T]] = ???

避免使用Await？

Answer 1

像这样:

def simplify[T](arg: Seq[Future[Option[Future[T]]]]):Future[Seq[T]] = {
  val seq: Future[Seq[Option[Future[T]]]] = Future.sequence(arg)
  val ret1 = seq.map { seq => Future.sequence(seq.flatten)}
  ret1.flatMap(identity)
}