JavaFX场景构建器访问不同类中的方法

Question

我一直在谷歌搜索和跟随不同的教程来学习如何做各种事情，到目前为止，我的进展还不错。但现在我已经停滞不前了。

我想做的是，当我的应用程序启动和用户按下登录按钮时，将它们转发到管理员主页上。

这是login.java中的代码：

public class Login extends Application {

private BorderPane quizApp;
private Stage windowLogin;

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

@Override
public void start(Stage windowLogin) {
    this.windowLogin = windowLogin;
    this.windowLogin.setTitle("QuizApp - Please Login or Register");

    windowLogin.resizableProperty().setValue(Boolean.FALSE);

    applyQuizAppLayout();
    applyLoginLayout();
}

public void applyQuizAppLayout() {
    try {
        FXMLLoader fxmlLoader = new FXMLLoader(Login.class.getClassLoader().getResource("view/QuizApp.fxml"));
        quizApp = (BorderPane) fxmlLoader.load();

        Scene scene = new Scene(quizApp);
        windowLogin.setScene(scene);
        windowLogin.show();
    } catch (IOException exception) {
        exception.printStackTrace();
    }
}

public void applyLoginLayout() {
    try {
        FXMLLoader fxmlLoader = new FXMLLoader(Login.class.getClassLoader().getResource("view/Login.fxml"));
        AnchorPane windowLogin = (AnchorPane) fxmlLoader.load();
        quizApp.setCenter(windowLogin);
    } catch (IOException exception) {
        exception.printStackTrace();
    }
}

public Stage getWindowLogin() {
    return windowLogin;
}

Button loginButton = new Button();

@FXML
private void handleLoginButtonAction(ActionEvent event) {
        System.out.println("Login details correct.");
        // windowLogin.setScene(AdministratorHomescreen.start(windowAdministratorHomescreen));
    }
}
}

public class AdministratorHomescreen {

private BorderPane quizApp;
private Stage windowAdministratorHomescreen;

public void start(Stage windowAdministratorHomescreen) {
    this.windowAdministratorHomescreen = windowAdministratorHomescreen;
    this.windowAdministratorHomescreen.setTitle("QuizApp - Welcome Administrator");

    windowAdministratorHomescreen.resizableProperty().setValue(Boolean.FALSE);

    applyQuizAppLayout();
    applyWindowAdministratorHomescreenLayout();
}

public void applyQuizAppLayout() {
    try {
        FXMLLoader fxmlLoader = new FXMLLoader(Login.class.getClassLoader().getResource("view/QuizApp.fxml"));
        quizApp = (BorderPane) fxmlLoader.load();

        Scene scene = new Scene(quizApp);
        windowAdministratorHomescreen.setScene(scene);
        windowAdministratorHomescreen.show();
    } catch (IOException exception) {
        exception.printStackTrace();
    }
}

public void applyWindowAdministratorHomescreenLayout() {
    try {
        FXMLLoader fxmlLoader = new FXMLLoader(
                Login.class.getClassLoader().getResource("view/WindowAdministratorHomescreen.fxml"));
        AnchorPane windowAdministratorHomescreen = (AnchorPane) fxmlLoader.load();
        quizApp.setCenter(windowAdministratorHomescreen);
    } catch (IOException exception) {
        exception.printStackTrace();
    }
}

public Stage getWindowAdministratorHomescreen() {
    return windowAdministratorHomescreen;
}
}

非常感谢任何帮助。

附注。当用户单击该按钮时，我不只是想打开任意场景，而是想打开一个已经应用了quizapp.fxml、quizapp.css和administratorlogin.fxml的场景，这就是我的代码布局。

Answer 1

在listener for log on按钮中，您需要将相关信息传递到下一个阶段。

首先在Logon类中为log in按钮设置一个侦听器

loginbutton.setOnAction(new EventHandler<ActionEvent>() {
   @Override
   public void handle(ActionEvent t) {
     try{
       if("the login is successful"){
         String temp = "Your relevant user information, like username or however you keep track of your data";
         windowLogin.hide();
         //You'll then make a new stage to pass the info to the administrator page
         AdministratorHomeScreen nextStage = new AdministratorHomeScreen(windowLogin, temp);
        }
        else{
              System.out.println("incorrect login);
            }
        }
catch(Exception e){
e.printStackTrace();
}
}
});

然后在AdministratorHomeScreen类中执行

public AdministratorHomeScreen(Stage stage, String takenOver){
windowLogIn = stage;
temp = takenOver;
start(new Stage());
}