如何从Scala中创建ScalaFX应用程序？

Question

I'm trying to launch the ScalaFX Hello World application from http://www.scalafx.org with the following code:

package car.cadr

object ApplicationStarter {
    def main(args: Array[String]) =
        javafx.application.Application.launch(classOf[HelloStageDemo], args: _*)
}

To clarify, I have two Scala files in the car.cadr package: ApplicationStarter.scala and HelloStageDemo.scala . HelloStageDemo.scala starts and runs perfectly fine, but the compiler is complaining about not found: type HelloStageDemo on ApplicationStarter.scala . Even if I manually import it with import car.cadr.HelloStageDemo the compiler still complains.

I'm using Scala 2.11.1 and ScalaFx 8.0.20-R6.

Answer 1

You have several problems here.

Let's start with the one the compiler is telling you about: not found: type HelloStageDemo . This makes sense because the HelloStageDemo example defines an object and not a class: so the scalac compiler actually outputs a class named HelloStageDemo$ (because you could also define a class HelloStageDemo , and both need to be compiled with different names).

---

Next, if you change your object HelloStageDemo for a class HelloStageDemo , you will get the following error:

Error:(7, 36) overloaded method value launch with alternatives:
  (x$1: String*)Unit <and>
  (x$1: Class[_ <: javafx.application.Application],x$2: String*)Unit
 cannot be applied to (Class[car.cadr.HelloStageDemo], String)

This is because the launch method exists only with the following signatures (here in Java):

• public static void launch(Class<? extends javafx.application.Application> appClass, String... args)
• public static void launch(String... args)

But HelloStageDemo is neither a String nor a kind of javafx.application.Application , so this cannot work.

---

This is because of the way ScalaFX's JFXApp trait works. Here's the main metrhod that gets executed when you launch a ScalaFX application the usual way (ie., the main class is the one extending JFXApp ):

import javafx.{application => jfxa}

trait JFXApp extends DelayedInit {
  // ...code removed for clarity...
  def main(args: Array[String]) {
    JFXApp.ACTIVE_APP = this
    arguments = args
    // Put any further non-essential initialization here.
    /* Launch the JFX application.
    */
    jfxa.Application.launch(classOf[AppHelper], args: _*)
  }
  // ...code removed for clarity...
}

So, in ScalaFX, the class extending javafx.application.Application isn't the one you implement, but a AppHelper class provided by ScalaFX. Notice that the main method first sets the ACTIVE_APP property on JFXApp 's companion object: in practice, what AppHelper will do is start JFXApp.ACTIVE_APP . Here is the code:

package scalafx.application

private[application] class AppHelper extends javafx.application.Application {
  def start(stage: javafx.stage.Stage) {
    JFXApp.STAGE = stage
    JFXApp.ACTIVE_APP.init()
    if (JFXApp.AUTO_SHOW) {
      JFXApp.STAGE.show()
    }
  }

  override def stop() {
    JFXApp.ACTIVE_APP.stopApp()
  }
}

---

In conclusion, if you want to launch HelloStageDemo but, for some reason, you don't want HelloStageDemo to be the main class, the simplest solution would be to just call the main method - after all, it's just a method like any other:

package car.cadr

object ApplicationStarter {
  def main(args: Array[String]) =
    HelloStageDemo.main(Array())
}

But if, for some reason, you absolutely had to launch your ScalaFX application trough the javafx.application.Application.launch method, I think the best solution would be to re-implement the AppHelper class to your liking, which seems like it should be pretty simple.

Answer 2

Here is a simple template for launching ScalaFX applications.

object MyApp {
   def main(args: Array[String]) {
        MyApp.launch(classOf[MyApp], args: _*)
    }
}

class MyApp extends JFXApp {

   override def start(primaryStage: Stage): Unit = {
      // initialization here
   } 
}

Answer 3

Non ScalaFX solution:

I know this question is answered and old. But if you want to use javafx.application.Application, to launch. Just set different names for the class and the object, with the result that there is no "MyClass$.class" and "MyClass.class" where the second one isn't the Applications Child but the object.

Solved my problem in a neat way.