如何在sbt子項目build.sbt文件中獲得對當前項目的引用

Question

我有一個playframework子項目。 目前，我將其定義為父build.sbt的playframework模塊。

lazy val silhouetteModule = (project in file("modules/silhouette"))
    .enablePlugins(PlayScala)

我傾向於將該知識委托給子項目build.sbt。 如何在子項目中調用此enablePlugins？

import play.PlayScala

scalaVersion := "2.11.1"

name := "play-silhouette-seed"

version := "1.0"

libraryDependencies ++= Seq(
  "com.mohiva" %% "play-silhouette" % "1.0",
  "org.webjars" %% "webjars-play" % "2.3.0",
  "org.webjars" % "bootstrap" % "3.1.1",
  "org.webjars" % "jquery" % "1.11.0",
  "net.codingwell" %% "scala-guice" % "4.0.0-beta4",
  cache
)

//this doesn't work
currentThisProject.enablePlugins(PlayScala)

Answer 1

您可以在子項目的build.sbt文件中直接調用enablePlugins ：

enablePlugins(PlayScala)