Question

请考虑以下代码段：

在此，我试图从未来使用＆＃39; For - yield＆＃39;理解。现在在yield方法中，我需要做一个检查，它调用函数fallbackResult，它返回一个future，因此getData的返回类型变为＆＃39; Future [Future [Option [Int]]]＆＃39;而不是未来[Option [Int]]＆＃39;。我怎么能以更好的方式做到这一点？（我确实使用了Map＆amp; FlatMap＆＃39; s，但由于嵌套了Maps和FlatMaps，代码有点难看）

def getData(): Future[Future[Option[Int]]] = {

  /* These are two future vectors. Ignore the Objects */

  val substanceTableF: Future[Vector[OverviewPageTableRowModel]] = getSubstanceTable(substanceIds, propertyId, dataRange)
  val mixtureTableF: Future[Vector[OverviewPageTableRowModel]] = getMixtureTableForSubstanceCombination(substanceIds, propertyId, dataRange)

  /* I have put for yeild to get values from futures.*/
  for {
      substanceTable <- substanceTableF
      mixtureTable <- mixtureTableF
  } yield {
      if(substanceTable.isEmpty && mixtureTable.isEmpty) {
          val resultF = fallbackResult()
          resultF.map(result => {Some(result)})
      } else {
          Future.successful(Some(100))
      }
  }
}

private def fallbackResult(): Future[Int] = { 
    // This method returns future of int
}

Answer 1

有很多方法可以解决这个问题，但关键是将yield逻辑移到for理解中。一种方法如下：

for {
  substanceTable <- substanceTableF
  mixtureTable <- mixtureTableF
  result <- (substanceTable.headOption orElse mixtureTable.headOption)
               .map(_ => Future.successful(Some(100)))
               .getOrElse(fallbackResult)
} yield result

Answer 2

我把代码放在for-corehension中：

for {
   substanceTable <- substanceTableF
   mixtureTable <- mixtureTableF 
   result <- {
     if (substanceTable.isEmpty && mixtureTable.isEmpty)
       fallbackResult()
     else
       Future.successful(Some(100))
   }
 } yield result

Scala：处理从Yeild返回Future的更好方法，以避免未来的未来

2 个答案: