scala - 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 中的一些代码。是否可以在该任务定义内从子项目 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. 

上面的代码会引发错误。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。

标签： scalaplayframeworksbtscala-reflect

Answer 1

来自 build.sbt 的参考 scala 文件

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")
}

然后在根build.sbt我可以打电话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"))

如果在 sbt shell 中运行root/sampleUnitTask它会打印foo.

---

我用sbt new playframework/play-scala-seed.g8. 一切似乎都奏效了。我添加了project/build.sbt和subproject1/src/main/scala/com/example/package1/App.scala如上。然后使用以下根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 prints 中执行foo。