Controller.java文件中的Javafx 2.0操作方法Application.getParameters()

Question

考虑以下样本.

如何访问控制器中应用程序的参数/参数？

谢谢.

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注意:我曾尝试将App.java和MyController.java混合在一个Class文件中,但没有帮助.

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App.java(简化):

public class App extends Application {

    public static void main(String[] args) {
        Application.launch(App.class, args);
    }

    @Override
    public void start(Stage primaryStage) throws Exception {

        // output arguments in console
        System.out.println(getParameters().getNamed().toString());

        Parent root = FXMLLoader.load(getClass().getResource("MyView.fxml"));
        final Scene scene = new javafx.scene.Scene(root);
        primaryStage.setScene(scene);
        primaryStage.show();
    }

}

MyController.java(简化):

public class MyController implements Initializable {

    @Override
    public void initialize(URL url, ResourceBundle rb) {
        // HOW TO getParameters() HERE ?
    }

    @FXML
    private Button myButton;

    @FXML
    private void my_Action(ActionEvent event) {
        // HOW TO getParameters() HERE ?
    }

}

MyView.fxml(简化):

<AnchorPane fx:id="root" fx:controller="mainpackage.MyController">
    <children>
        <Button fx:id="myButton" onAction="#my_Action" text="Start" />
    </children>
</AnchorPane>

Answer 1

1.最直接的方式 - 将它们保存在应用程序中:

public class App extends Application {

  public static void main(String[] args) { launch(); }

  public static String parameters;

  @Override
  public void start(Stage primaryStage) throws Exception {

    parameters = getParameters().getNamed().toString();

    Parent root = FXMLLoader.load(getClass().getResource("MyView.fxml"));
    final Scene scene = new javafx.scene.Scene(root);
    primaryStage.setScene(scene);
    primaryStage.show();
  }

}

并在控制器中读取它们:

public class MyController implements Initializable {

  @Override
  public void initialize(URL url, ResourceBundle rb) {
     System.out.println(App.parameters);
  }

2.下一个主题中描述了更复杂(但总体上更好)的方法:

• 传递参数JavaFX FXML
• 带控制器的多个FXML,共享对象