Question

I have cut my problem down to simple lines of Scala + JavaFX 2’s FXMLLoader:

val path = "/com/myapp/views/main.fxml"

val loader = new FXMLLoader()
loader.setLocation(getClass.getResource(path))

val root = loader.load(getClass.getResourceAsStream(path)).asInstanceOf[Parent]

I am trying to load a main.fxml file using FXMLLoader, but I end up with:

Class javafx.fxml.FXMLLoader$ValueElement can not access a member of
class com.myapp.controllers.MainWindow with modifiers “private”

The FXML code looks like:

<?xml version="1.0" encoding="UTF-8"?>

<?import java.lang.*?>
<?import java.net.*?>
<?import java.util.*?>
<?import javafx.collections.*?>
<?import javafx.geometry.*?>
<?import javafx.scene.Scene?>
<?import javafx.scene.control.*?>
<?import javafx.scene.effect.*?>
<?import javafx.scene.layout.*?>
<?import javafx.scene.paint.*?>
<?import javafx.scene.text.*?>

<BorderPane fx:controller="com.myapp.controllers.MainWindow" fx:id="mainWindow" prefHeight="703.0" prefWidth="803.0" xmlns:fx="http://javafx.com/fxml">
  <stylesheets>
    <URL value="@../styles/Styles.css" />
  </stylesheets>
  <top>
  ...

According to the error message, JavaFX’s FXMLLoader tries to access some property of my controller which is private. However, I have no private members in the controller:

class MainWindow extends Initializable {
  override def initialize(location: URL, resourceBundle: java.util.ResourceBundle) {
    print("init")
  }
}

What could be the problem?

Answer 1

The error points to a private constructor. Do you need to explicitly make a constructor public in Scala? You can use javap to see what is compiled.