Question

我是JavaFX的新手，在做项目时，我试图切换屏幕。我正在考虑将相应的AnchorPane可见性设置为true或false，并且从另一个Controller访问AnchorPane时遇到问题。我试图使AnchorPane静态，但它提供了NullPointerException。

这是我的代码。

控制器类

public class Controller {

    @FXML
    AnchorPane signInPane;

    @FXML
    private TextField usernameForSignIn;

    @FXML
    private PasswordField password;

    @FXML
    private Button signIn;

    @FXML
    private Button registration;

    @FXML
    void initialize() {
        registration.setOnAction(event -> {
            signInPane.setVisible(false);
            SignUpController.registerPane.setVisible(true);
        });

        signIn.setOnAction(event ->  {
            String usernameText = usernameForSignIn.getText().trim();
            String passwordText = password.getText().trim();

            if(!usernameText.equals("") && !passwordText.equals("")) {
                loginUser(usernameText, passwordText);
            } else {
                System.out.println("Empty login and/or password");
            }

        });
    }

    private void loginUser(String usernameText, String passwordText) {
    }

}

SignUpController类

public class SignUpController {

    @FXML
    static AnchorPane registerPane;

    @FXML
    private TextField email;

    @FXML
    private PasswordField pass;

    @FXML
    private Button signUp;

    @FXML
    private TextField fname;

    @FXML
    private TextField lname;

    @FXML
    private TextField username;

    @FXML
    private RadioButton radioMale;

    @FXML
    void initialize() {

        signUp.setOnAction(event -> {

            signUpNewUser();

        });
    }

    private void signUpNewUser() {
        DatabaseHandler databaseHandler = new DatabaseHandler();

        String firstName = fname.getText();
        String lastName = lname.getText();
        String usname = username.getText();
        String password = pass.getText();
        String e_mail =  email.getText();
        String gender = "";
        if(radioMale.isSelected()) {
            gender = "Male";
        } else {
            gender = "Female";
        }

        User user = new User(firstName,lastName,usname,password,e_mail,gender);

        databaseHandler.signUpUser(user);

    }
}

我需要从此屏幕转到

对此，无需关闭窗口

我试图解决这样的问题，但是它关闭了窗口并打开了一个新窗口。

registration.setOnAction(event -> {
    registration.getScene().getWindow().hide();

    FXMLLoader loader = new FXMLLoader();
    loader.setLocation(getClass().getResource("/sample/view/signUp.fxml"));

    try {
        loader.load();
    } catch (IOException e) {
        e.printStackTrace();
    }

    Parent root = loader.getRoot();
    Stage stage = new Stage();
    stage.setScene(new Scene(root));
    stage.showAndWait();

// signInPane.setVisible(false);
});

Answer 1

创建一个窗口并重复使用，而不是每次都创建一个新窗口。然后，您可以只检查它是否显示，也可以禁用显示时显示的控件。

Answer 2

类似这样的方法将确保在initalize（）之前加载FXML。您不必隐藏您的AnchorPane，因为实例已经加载。隐藏从FXML创建的项目通常会导致错误。

public class Controller {

    @FXML
    Button button_CustomerInfo;
    @FXML
    Button button_A1Child;
    @FXML
    BorderPane bPane;
    @FXML
    ToggleGroup tog;
    Parent r1;
    Parent r2;

    // CREATE SOMETHING LIKE THIS
    // THIS SAMPLE CODE USES THE SAME BORDERPANE
    // AND THE CENTER OF THE BORDERPANE
    // IS CHANGED WHEN A BUTTON IS PRESSED
    {
        try {
            r1 = FXMLLoader.load(getClass().getResource("SecondSreen.fxml"));
            r2 = FXMLLoader.load(getClass().getResource("FirstScreen.fxml"));
        } catch (IOException e) {
            e.printStackTrace();
        }


    }

    public void initialize() {


        button_CustomerInfo.setOnAction(e-> {
            bPane.setCenter(r2);
        });

        button_A1Child.setOnAction(e-> {
            bPane.setCenter(r1);
        });
    }
}

切换到不同的屏幕JavaFX和FXML

2 个答案: