Sbt：从根项目调用子项目的源代码

Question

所以我的项目结构如下所示：我的根项目包含设置、任务和配置。此外还有一个包含所有子项目的子项目文件夹。我在根项目上创建了一个任务，该任务依赖于子项目 X 中的某些代码。是否可以从该任务定义中的子项目 x 调用方法？

我的代码如下所示：

lazy val rootSettings: Seq[Setting[_]] = Seq (someRootTask :=  { //I need to call an object from a subproject here..}) 

我尝试使用反射api没有成功：

import scala.reflect.runtime.{universe => u }
lazy val docSettings: Seq[Setting[_]] = Seq(
  rootTask := {
      val subproject = baseDirectory.in(playApp).value.getAbsolutePath
      val mirror = u.runtimeMirror(getClass.getClassLoader)
      val clazz = mirror.staticModule(subproject+"/" +"controllers.pckg.obj" ) 

      val cm = mirror.reflectModule(clazz)
      val instanceMirror = mirror.reflect(cm.instance)
      val methodName ="sayHi" 

      val methodSymbol = u.typeOf[instanceMirror.type].declaration(u.newTermName(methodName)).asMethod
      val method = instanceMirror.reflectMethod(methodSymbol)

      method.apply()
  }
)
// still can't point to the object i want to call. 

上面的代码抛出一个错误。它找不到对象，我知道它的路径，但我不能从根项目中将它作为 package.class 引用。

Answer 1

Reference scala file from build.sbt

如果我理解正确，你的项目看起来像

root
  project
    build.properties
    build.sbt
  subproject1
    src
      main
        scala
          com.example.package1
            App.scala
  subproject2
    src
      main
        scala
          com.example.package2
  build.sbt

在project/build.sbt我可以写

Compile / unmanagedSourceDirectories += baseDirectory.value / ".." / "subproject1" / "src" / "main" / "scala"

假设subproject1/src/main/scala/com/example/package1/App.scala是

package com.example.package1

object App {
  def foo(): Unit = println("foo")
}

然后在rootbuild.sbt我可以调用foo

name := "sbtdemo"

version := "0.1"

ThisBuild / scalaVersion := "2.13.4"

lazy val sampleUnitTask = taskKey[Unit]("A sample unit task.")

lazy val rootSettings: Seq[Setting[_]] = Seq(
  sampleUnitTask := {
    com.example.package1.App.foo()
  }
)

lazy val root = project
  .in(file("."))
  .dependsOn(subproject1, subproject2)
  .settings(rootSettings)

lazy val subproject1 = project
  .in(file("subproject1"))

lazy val subproject2 = project
  .in(file("subproject2"))

如果我在 sbt shell 中运行root/sampleUnitTask，它会打印出foo。

---

我使用sbt new playframework/play-scala-seed.g8 创建了 Play 项目。一切似乎都奏效了。我如上所述添加了project/build.sbt 和subproject1/src/main/scala/com/example/package1/App.scala。然后用下面的根build.sbt

name := """playframeworkdemo"""
organization := "com.example"

version := "1.0-SNAPSHOT"

lazy val sampleUnitTask = taskKey[Unit]("A sample unit task.")

lazy val rootSettings: Seq[Setting[_]] = Seq(
  sampleUnitTask := {
    com.example.package1.App.foo()
  }
)

lazy val root = (project in file(".")).enablePlugins(PlayScala)
  .dependsOn(subproject1, subproject2)
  .settings(rootSettings)

ThisBuild / scalaVersion := "2.13.3"

libraryDependencies += guice
libraryDependencies += "org.scalatestplus.play" %% "scalatestplus-play" % "5.0.0" % Test

// Adds additional packages into Twirl
//TwirlKeys.templateImports += "com.example.controllers._"

// Adds additional packages into conf/routes
// play.sbt.routes.RoutesKeys.routesImport += "com.example.binders._"

lazy val subproject1 = project
  .in(file("subproject1"))

lazy val subproject2 = project
  .in(file("subproject2"))

root/sampleUnitTask 在 sbt shell 中执行打印 foo。