如何将 sbt 0.12 "<<= map case" 转换为 sbt 1.x

Question

sbt 0.12 build.sbt

val myTask = TaskKey[Unit]("mytask")
lazy val baseSettings = Seq(
    javacOptions ++= Seq("-g", "-encoding", "utf8")
)
lazy val appSettings = baseSettings ++ Seq(
  myTask <<= (fullClasspath in Runtime, normalizedName, target, packageOptions, streams, cacheDirectory) map {
    case (fullClasspath: Classpath, normalizedName: String, target: File, packageOptions: Seq[PackageOption], streams: TaskStreams, cacheDirectory: File) => {
      streams.log.info("my task appSettings")
    }
  }
)

想将以下代码更新为sbt 1.x

myTask <<= (fullClasspath in Runtime, normalizedName, target, packageOptions, streams, cacheDirectory) map {
    case (fullClasspath: Classpath, normalizedName: String, target: File, packageOptions: Seq[PackageOption], streams: TaskStreams, cacheDirectory: File) => {

谢谢

Update:
I find some information at scala-sbt.org/0.13/docs/Migrating-from-sbt-012x.html, but I don't know how to use ":=" with "case"

EDIT:
Mateusz Kubuszok is right, not need "case", and "cacheDirectory" can be replaced by "streams.value.cacheDirectory"

Answer 1

而不是 <<= 你使用 :=.

而不是设置元组和 map，您只需在 key/settings 上调用 .value。

cacheDirectory 已删除。

从 1.1 开始还有 unified path syntax

lazy val appSettings = Seq(
  myTask := {
    streams.value.log.info(s"my task appSettings: ${(Runtime / fullClasspath).value}, ${normalizedName.value}, ${target.value}, ${packageOptions.value}")
  }
)

在更复杂的任务中，您可能需要查阅有关例如order of execution.

如何将 sbt 0.12 "<<= map case" 转换为 sbt 1.x

How to convert sbt 0.12 "<<= map case" to sbt 1.x

scala

sbt