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scala - 如何将源映射到另一个？

问题描述

我有两个案例类

case class Color (name: String, shades: List[Shade] = List.empty)

case class Shade (shadeName: String)

我也有两个解析器：

object ColorParser {
    def apply(
    s: String): Either[List[SomethingElse], Color] = {
        val values = s.split("\\|", -1).map(_.trim).toList
        validateColor(values).leftMap(xs => xs.toList).toEither
    }
}

object ShadesParser {
    def apply(s: String)
    : Either[List[SomethingElse], Shade] = {
        val values = s.split('|').map(_.trim).toList
        validateShade(values).leftMap(xs => xs.toList).toEither
    }
}

我有一个来源Color和一个来源Shade。

  sourceForShade
  .via (framing("\n"))
  .map (_.utf8string)
  .map (_.trim)
  .map {
    s => ShadesParser(s)
  }
  .collect {
    case Right(shade) => shade
  }

  sourceForColor
  .via(framing("\n"))
  .map(_.utf8String)
  .map(_.trim)
  .map(s => ColorParser(s))
  .collect {
    case Right(color) => color
  }
  .map {color =>
       //Here I want access to Color object that has the property shades list property set based on sourceForShade. 
       //At the moment it only has the name field but the List[Shade] property is empty.
  }

问题

在 的评论部分map，我如何才能访问一个颜色对象，该对象也具有shades: List[Shade]基于填充的属性sourceForShade

标签： scalaakkaakka-stream

解决方案

一种方法是首先Future[Seq[Shade]]从第一个流中获取 a，然后将其结果应用于Future第二个流：

val shades: Future[Seq[Shade]] =
  sourceForShade
    .via(framing("\n"))
    .map(_.utf8String.trim)
    .map(ShadesParser)
    .collect {
      case Right(shade) => shade
    }
    .runWith(Sink.seq[Shade])

val colors: Future[Source[Color, _]] =
  shades map { s =>
    sourceForColor
      .via(framing("\n"))
      .map(_.utf8String.trim)
      .map(ColorParser)
      .collect {
        case Right(color) => color
      }
      .map(c => c.copy(shades = s.toList))
  }

val colorsWithShades: Source[Color, _] =
  Source.fromFuture(colors).flatMapConcat(identity)

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