How to swap screens in a javafx-application in the

2019-01-18 03:56发布

Hey I searched the net for quite a while but I couldn't find a solution to the following problem:

In javafx you got 3 basic files; the controller-class, the fxml file and the application class. Now I want to react in the controller to a button-click (which works perfectly fine) and change the screen on that click (which you normally do with stage.setScreen() ), but I have no reference to the stage (which you can find in the application class).

Application-Sample:

public class JavaFXApplication4 extends Application {

@Override
public void start(Stage stage) throws Exception {
    Parent root = FXMLLoader.load(getClass().getResource("Sample.fxml"));

    Scene scene = new Scene(root);

    stage.setScene(scene);
    stage.show();
}

/**
 * The main() method is ignored in correctly deployed JavaFX application.
 * main() serves only as fallback in case the application can not be
 * launched through deployment artifacts, e.g., in IDEs with limited FX
 * support. NetBeans ignores main().
 *
 * @param args the command line arguments
 */
public static void main(String[] args) {
    launch(args);
}
}

FXML-Sample:

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

<?import java.lang.*?>
<?import java.util.*?>
<?import javafx.scene.*?>
<?import javafx.scene.control.*?>
<?import javafx.scene.layout.*?>

<AnchorPane id="AnchorPane" prefHeight="200.0" prefWidth="320.0"  xmlns:fx="http://javafx.com/fxml" fx:controller="javafxapplication4.SampleController">
  <children>
  <Button id="button" fx:id="nextScreen" layoutX="126.0" layoutY="90.0" onAction="#handleButtonAction" text="Next Screen" />
  <Label fx:id="label" layoutX="126.0" layoutY="120.0" minHeight="16.0" minWidth="69.0" />
  </children>
</AnchorPane>

Controller-Sample:

public class SampleController implements Initializable {

@FXML
private Label label;

@FXML
private void handleButtonAction(ActionEvent event) {
    System.out.println("You clicked me!");
    label.setText("Hello World!");
    //Here I want to swap the screen!
}

@Override
public void initialize(URL url, ResourceBundle rb) {
    // TODO
}    
}

I would be thankful for any kind of help.

3条回答
Ridiculous、
2楼-- · 2019-01-18 04:02
@FXML
private void handleButtonAction(ActionEvent event) {
    System.out.println("You clicked me!");
    label.setText("Hello World!");
    //Here I want to swap the screen!

    Stage stageTheEventSourceNodeBelongs = (Stage) ((Node)event.getSource()).getScene().getWindow();
    // OR
    Stage stageTheLabelBelongs = (Stage) label.getScene().getWindow();
    // these two of them return the same stage
    // Swap screen
    stage.setScene(new Scene(new Pane()));
}
查看更多
兄弟一词,经得起流年.
3楼-- · 2019-01-18 04:08

I found this old question while getting into Java and trying to solve the same thing. Since I wanted the scenes to remember the content between the switches, I couldn't use the accepted answer, because when switching between the scenes it instantiates them again (loosing their previous state).

Anyways the accepted answer and the answer to the similar question gave me a hints on how to switch the scenes without loosing their states. The main idea is to inject instance of the scene into another controller, so that the controller doesn't need to instantiate a fresh scene over and over again but can use already existing instance (with its state).

So here is the main class that instantiates the scenes:

public class Main extends Application {

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

    @Override
    public void start(Stage primaryStage) throws Exception {
        // getting loader and a pane for the first scene. 
        // loader will then give a possibility to get related controller
        FXMLLoader firstPaneLoader = new FXMLLoader(getClass().getResource("firstLayout.fxml"));
        Parent firstPane = firstPaneLoader.load();
        Scene firstScene = new Scene(firstPane, 300, 275);

        // getting loader and a pane for the second scene
        FXMLLoader secondPageLoader = new FXMLLoader(getClass().getResource("secondLayout.fxml"));
        Parent secondPane = secondPageLoader.load();
        Scene secondScene = new Scene(secondPane, 300, 275);

        // injecting second scene into the controller of the first scene
        FirstController firstPaneController = (FirstController) firstPaneLoader.getController();
        firstPaneController.setSecondScene(secondScene);

        // injecting first scene into the controller of the second scene
        SecondController secondPaneController = (SecondController) secondPageLoader.getController();
        secondPaneController.setFirstScene(firstScene);

        primaryStage.setTitle("Switching scenes");
        primaryStage.setScene(firstScene);
        primaryStage.show();
    }
}

And here are the both controllers:

public class FirstController {

    private Scene secondScene;

    public void setSecondScene(Scene scene) {
        secondScene = scene;
    }

    public void openSecondScene(ActionEvent actionEvent) {
        Stage primaryStage = (Stage)((Node)actionEvent.getSource()).getScene().getWindow();
        primaryStage.setScene(secondScene);
    }
}

yep, second one looks the same (some logic could probably be shared, but the current state is enough as a proof of concept)

public class SecondController {

    private Scene firstScene;

    public void setFirstScene(Scene scene) {
        firstScene = scene;
    }

    public void openFirstScene(ActionEvent actionEvent) {    
        Stage primaryStage = (Stage)((Node)actionEvent.getSource()).getScene().getWindow();
        primaryStage.setScene(firstScene);
    }
}
查看更多
淡お忘
4楼-- · 2019-01-18 04:08

You can try like this too.

public void onBtnClick(ActionEvent event) {
    try {
        FXMLLoader loader = new FXMLLoader(getClass().getResource("login.fxml"));
        Stage stage = (Stage) btn.getScene().getWindow();
        Scene scene = new Scene(loader.load());
        stage.setScene(scene);
    }catch (IOException io){
        io.printStackTrace();
    }

}
查看更多
登录 后发表回答