Play框架和Scala.tools.nsc

Question

import scala.tools.nsc._

trait Foo
class Parser {
    def parse(code: String) = {
        val settings = new Settings
        settings.embeddedDefaults[Foo]
        val interpreter = new Interpreter(settings)
        interpreter.parse(code)
    }
}

"org.scala-lang" % "scala-compiler" % "2.9.1"

**编程对象，Settings.UseJavaCP.Value=true。

Build.Scala

import sbt._
import Keys._
import PlayProject._

object ApplicationBuild extends Build {

    val appName         = "com.qwerty.utils"
    val appVersion      = "1.0-SNAPSHOT"
    val scalaVersion    = "2.9.1"

    val appDependencies = Seq(
        "org.scala-lang" % "scala-compiler" % "2.9.1"
    )
    
    val main = PlayProject(appName, appVersion, appDependencies, mainLang = SCALA).settings(
      // Add your own project settings here
    
    )

}

Answer 1

有关embeddeddefaults的背景信息，请参阅原始建议。

容器（Play）必须定义“app.class.path”和“boot.class.path”资源，然后embeddeddefaults将使用它们为环境正确配置解释器。所以，这是一个游戏的增强。

如果您可以将必要的类路径传递到应用程序中，那么您可以自己显式地配置类路径和类加载器，如下所示：

val settings = new Settings
settings.classpath.value = "<classpath>"
settings.bootclasspath.value =
  settings.bootclasspath.value + File.pathSeparator +
  "<extra-bootclasspath>"
val interpreter = new Interpreter(settings) {
    override def parentClassLoader = classOf[Foo].getClassLoader
}
interpreter.parse(code)