Question

如何以编程方式（在build.sbt中）在sbt 0.13中找到当前根项目的所有子项目？（我尚未尝试过Project.componentProjects，因为它是sbt 1.0中的新功能。）

lazy val root = (project in file(".") ... )

val myTask = taskKey[Unit]("some description")

myTask := {
  val masterRoot = baseDirectory.value
  // This does not work
  // val subProjects: Seq[ProjectReference] = root.aggregate
  // So I tried to specify the subproject list explicitly; still does not work
  val subProjects = Seq[Project](subPrj1)
  subProjects.foreach { subproject =>
    // All of this works if the "subproject" is hard-coded to "subPrj1"
    val subprojectTarget = target.in(subproject).value / "classes"
    val cp = (dependencyClasspath in(subproject, Compile, compile)).value
  }
}

得到这些错误：

build.sbt: error: Illegal dynamic reference: subproject
    val subprojectTarget = target.in(subproject).value / "classes"
                                     ^
build.sbt: error: Illegal dynamic reference: subproject
    val cp = (dependencyClasspath in(subproject, Compile, compile)).value

Answer 1

您可以通过buildStructure.value.allProjectRefs访问所有子项目的列表。

另一部分是您的问题，这是我也经常遇到的可怕问题。通过首先创建List[Task[A]，然后使用递归函数将其提升为Task[List[A]]，我能够解决这些问题。

def flattenTasks[A](tasks: Seq[Def.Initialize[Task[A]]]): Def.Initialize[Task[List[A]]] =
  tasks.toList match {
    case Nil => Def.task { Nil }
    case x :: xs => Def.taskDyn {
      flattenTasks(xs) map (x.value :: _)
    }
  }

myTask := {
  val classDirectories: List[File] = Def.taskDyn {
    flattenTasks {
      for (project ← buildStructure.value.allProjectRefs)
        yield Def.task { (target in project).value / "classes" }
    }
  }.value
}

我使用了这种方法，例如此处：utility methods actual usage

如何在SBT 0.13中动态获取子项目列表

1 个答案: