scala - 在 SBT 0.13 中习惯性地定义动态任务？

Question

我正在从 0.12 移动 SBT 插件转至 0.13 .在我的插件的各个点上，我将一组动态任务安排到 SBT 构建图上。

下面是我的旧代码。这仍然是表达这一点的惯用方式，还是可以利用宏使一切变得更漂亮？

import sbt._
import Keys._

object Toplevel extends Build
{
    lazy val ordinals = taskKey[Seq[String]]("A list of things")
    lazy val times = taskKey[Int]("Number of times to list things")
    lazy val inParallel = taskKey[Seq[String]]("Strings to log in parallel")

    lazy val Foo = Project( id="Foo", base=file("foo"),
        settings = Defaults.defaultSettings ++ Seq(
            scalaVersion    := "2.10.2",

            ordinals := Seq( "First", "Second", "Third", "Four", "Five" ),

            times := 3,

            inParallel <<= (times, ordinals, streams) flatMap
            { case (t, os, s) =>
                os.map( o => toTask( () =>
                {
                    (0 until t).map( _ => o ).mkString(",")
                } ) ).join
            }

        )
    )
}

为这个完全人为的例子道歉!

编辑

因此，考虑到马克的建议，我有以下更整洁的代码:

import sbt._
import Keys._

object Toplevel extends Build
{
    lazy val ordinals = taskKey[Seq[String]]("A list of things")
    lazy val times = taskKey[Int]("Number of times to list things")
    lazy val inParallel = taskKey[Seq[String]]("Strings to log in parallel")

    def parTask = Def.taskDyn
    {
        val t = times.value
        ordinals.value.map(o => ordinalTask(o, t)).join
    }

    def ordinalTask(o: String, t: Int) = Def.task
    {
        (0 until t).map(_ => o).mkString(",")
    }

    lazy val Foo = Project( id="Foo", base=file("foo"),
        settings = Defaults.defaultSettings ++ Seq(
            scalaVersion    := "2.10.2",

            ordinals := Seq( "First", "Second", "Third", "Four", "Five" ),

            times := 3,

            inParallel := parTask.value
        )
    )
}

这似乎就在那里，但构建失败:

[error] /home/alex.wilson/tmp/sbt0.13/project/build.scala:13: type mismatch;
[error]  found   : sbt.Def.Initialize[Seq[sbt.Task[String]]]
[error]  required: sbt.Def.Initialize[sbt.Task[?]]
[error]         ordinals.value.map(o => ordinalTask(o, t)).join

Answer 1

您可以使用 Def.taskDyn ，它为 flatMap 提供了新的语法.与Def.task的区别是预期的返回类型是任务 Initialize[Task[T]]而不仅仅是 T .翻译你的例子，

inParallel := parTask.value

def parTask = Def.taskDyn {
   val t = times.value
   ordinals.value.map(o => ordinalTask(o, t)).joinWith(_.join)
}

def ordinalTask(o: String, t: Int) = Def.task {
   (0 until t).map(_ => o).mkString(",")
}