Question

我的基本加密器应用程序出了问题。如果有人在 keyTextField 中键入字符串，我想生成一个错误窗口。还有一个使用OK按钮关闭错误窗口的事件（窗口图形从fxml文件加载）我尝试过如下所示，但是没有成功，我也使用了close（）方法。处理应用程序控制的最佳方法是什么？我只使用MainController，我认为这不是一个好主意。提前谢谢

`package pl.gumisok.cipherController;

import java.io.IOException;
import java.net.URL;
import java.util.ResourceBundle;

import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.fxml.FXML;
import javafx.fxml.FXMLLoader;
import javafx.fxml.Initializable;
import javafx.scene.Node;
import javafx.scene.Parent;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.TextArea;
import javafx.scene.control.TextField;
import javafx.stage.Stage;
import pl.gumisok.cipherMain.CipherManager;

public class MainController implements Initializable {

CipherManager cipher;
@FXML
private ContentPaneController contentPaneController;

@FXML
private ControlPaneController controlPaneController;

@Override
public void initialize(URL arg0, ResourceBundle arg1) {
    // TODO Auto-generated method stub

    System.out.println(contentPaneController);
    System.out.println(controlPaneController);

    Button encryptButton = controlPaneController.getEncryptButton();
    Button decryptButton = controlPaneController.getDecryptButton();
    Button okButton = controlPaneController.getOkButton();
    TextArea cleanTextArea = contentPaneController.getCleanTextArea();
    TextArea cryptTextArea = contentPaneController.getCryptTextArea();
    TextField keyTextField = controlPaneController.getKeyTextField();

    encryptButton.setOnAction(new EventHandler<ActionEvent>() {

        @Override
        public void handle(ActionEvent event) {
            String wiadomosc = cleanTextArea.getText();
            System.out.println(wiadomosc);
            try {
                int key = Integer.parseInt(keyTextField.getText());
                System.out.println(key);
            } catch (NumberFormatException e) {
                System.out.println(e);
                FXMLLoader fxmlLoader = new FXMLLoader(getClass()
                        .getClassLoader().getResource(
                                "pl/gumisok/cipherView/Error.fxml"));
                Parent root;
                try {
                    root = fxmlLoader.load();

                    Stage sstage = new Stage();

                    sstage.setOpacity(1);
                    sstage.setTitle("Error");
                    sstage.setScene(new Scene(root));
                    sstage.show();
                   okButton.setOnAction(x->sstage.hide());
                } catch (IOException e1) {
                    // TODO Auto-generated catch block
                    e1.printStackTrace();
                }
            }

        }

    });
}

}`

Answer 1

我希望，我理解这个问题是正确的，这是一个如何创建警报对话框的示例

Alert alert = new Alert(AlertType.INFORMATION);
alert.setTitle("Information Dialog");
alert.setHeaderText(null);
alert.setContentText("I have a great message for you!");
alert.showAndWait();

Answer 2

你的申请层不好您需要将fxml文件中的按钮操作绑定到控制器也许是这样的：

Error.fxml：

<AnchorPane xmlns="http://javafx.com/javafx/8"
        xmlns:fx="http://javafx.com/fxml/1"
        fx:controller="controllers.ErrorController">
    <children>
        <Label text="ERROR!" />
        <Button text="close" onAction="#hide" layoutY="15"/>
    </children>
</AnchorPane>

ErrorController.java：

public class ErrorController {

    private static Stage stage;
    private static Parent root;

    public ErrorController(){}
    public ErrorController(Window owner) throws IOException {
        if (root == null)
            root = FXMLLoader.load(ClassLoader
                    .getSystemResource("views/Error.fxml"));
        if (stage == null)
            stage = new Stage();
        //stage.initModality(Modality.WINDOW_MODAL);
        stage.initOwner(owner);
        stage.setTitle("Error");
        stage.setScene(new Scene(root));
    }
    public void show() {
        stage.show();
    }
    public @FXML void hide() {
        stage.hide();
    }
}

然后使用它

...
error = new ErrorController(node.getScene().getWindow());
...

try {
    int key = Integer.parseInt(keyTextField.getText());
    System.out.println(key);
} catch (NumberFormatException e) {
    error.show();    
}

JavaFX，使用Button关闭错误窗口

2 个答案: