播放子项目:如何转换为build.sbt

Question

我有一个工作的多模块Play 2.2应用程序,它的组织方式......

myApp
  + app
  + conf
  + project
      + build.properties
      + Build.scala
      + plugin.sbt

......其中Build.scala包含以下声明:

import sbt._
import Keys._
import play.Project._

object ApplicationBuild extends Build {

  val appName         = "myApp"
  val appVersion      = "1.0-SNAPSHOT"

  val authDependencies = Seq(
    "se.radley" %% "play-plugins-salat" % "1.3.0"
  )

  val mainDependencies = Seq(
    "se.radley" %% "play-plugins-salat" % "1.3.0"
  )

  lazy val auth = play.Project(
    appName + "-auth",
    appVersion,
    authDependencies,
    path = file("modules/auth")).settings(
      lessEntryPoints <<= baseDirectory(customLessEntryPoints),
      routesImport += "se.radley.plugin.salat.Binders._",
      templatesImport += "org.bson.types.ObjectId",
      testOptions in Test := Nil,
      resolvers ++= Seq(Resolvers.sonatype, Resolvers.scalaSbt)
    )

  lazy val main = play.Project(
    appName,
    appVersion,
    mainDependencies).settings(
      scalacOptions += "-language:reflectiveCalls",
      routesImport += "se.radley.plugin.salat.Binders._",
      templatesImport += "org.bson.types.ObjectId",
      testOptions in Test := Nil,
      lessEntryPoints <<= baseDirectory(customLessEntryPoints),
      resolvers ++= Seq(Resolvers.sonatype, Resolvers.scalaSbt)
    ).dependsOn(auth).aggregate(auth)

  def customLessEntryPoints(base: File): PathFinder = {
    (base / "app" / "assets" / "stylesheets" / "bootstrap" * "bootstrap.less") +++
    (base / "app" / "assets" / "stylesheets" * "*.less")
  }
}

object Resolvers {

  val scalaSbt = Resolver.url("Scala Sbt", url("http://repo.scala-sbt.org/scalasbt/sbt-plugin-snapshots"))(Resolver.ivyStylePatterns)
  val sonatype = Resolver.sonatypeRepo("snapshots")
}

现在阅读Play 2.2文档,看起来我应该将我的项目转换为build.sbt:

以下示例使用build.scala文件声明play.Project.这种方法是在2.2版之前定义Play应用程序的方式.保留该方法以支持向后兼容性.我们建议您转换为基于build.sbt的方法,或者,如果使用build.scala,则使用sbt的Project类型和项目宏.

是否有介绍如何更换任何工作示例project/build.scala用build.sbt？我在这里和那里读了一些简短的文章......但我无法得到一个有效的Play项目.

Answer 1

没有迫切需要将您的构建转换为build.sbt. build.sbt更简单,但基本上只是编译成Build.scala.

这个问题的另一个答案是可行的,但也许有点冗长.从SBT文档开始:

http://www.scala-sbt.org/0.13.0/docs/Getting-Started/Multi-Project.html

现在,创建指定主项目和子项目,并将主项目设置放入主build.sbt文件中:

lazy val auth = project.in(file("modules/auth"))

lazy val main = project.in(file(".")).dependsOn(auth).aggregate(auth)

playScalaSettings

name := "myApp"

version := "1.0-SNAPSHOT"

libraryDependencies += "se.radley" %% "play-plugins-salat" % "1.3.0"

scalacOptions += "-language:reflectiveCalls"

routesImport += "se.radley.plugin.salat.Binders._"

templatesImport += "org.bson.types.ObjectId"

testOptions in Test := Nil

lessEntryPoints <<= baseDirectory(customLessEntryPoints)

resolvers ++= Seq(Resolvers.sonatype, Resolvers.scalaSbt)

object Resolvers {
  val scalaSbt = Resolver.url("Scala Sbt", url("http://repo.scala-sbt.org/scalasbt/sbt-plugin-snapshots"))(Resolver.ivyStylePatterns)
  val sonatype = Resolver.sonatypeRepo("snapshots")
}

现在,在modules/auth/build.sbt,为auth模块设置您的设置:

name := "myApp-auth"

lessEntryPoints <<= baseDirectory(customLessEntryPoints)

routesImport += "se.radley.plugin.salat.Binders._"

templatesImport += "org.bson.types.ObjectId"

testOptions in Test := Nil

resolvers ++= Seq(Resolvers.sonatype, Resolvers.scalaSbt)

无论如何,它可能需要一些调整,但希望你明白这一点.