FXML文件中的getHostServicesshowDocument

如何解决FXML文件中的getHostServicesshowDocument

您需要将传递HostServices给Controller。

密码：在HostServices中设置Controller。

HostServices hostServices ;

public void setGetHostController(HostServices hostServices)
{
    this.hostServices = hostServices;
}

密码：传递HostServices给Controller。

FXMLLoader loader = new FXMLLoader(getClass().getResource("FXMLDocument.fxml"));
Parent root = loader.load();
FXMLDocumentController fXMLDocumentController = loader.getController();
fXMLDocumentController.setGetHostController(getHostServices());

主要

import javafx.application.Application;
import javafx.fxml.FXMLLoader;
import javafx.scene.Parent;
import javafx.scene.Scene;
import javafx.stage.Stage;

/**
 *
 * @author sedrick
 */
public class JavaFXApplication7 extends Application {

    @Override
    public void start(Stage stage) throws Exception {
        FXMLLoader loader = new FXMLLoader(getClass().getResource("FXMLDocument.fxml"));
        Parent root = loader.load();
        FXMLDocumentController fXMLDocumentController = loader.getController();
        fXMLDocumentController.setGetHostController(getHostServices());

        Scene scene = new Scene(root);        
        stage.setScene(scene);
        stage.show();
    }

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        launch(args);
    }

}

控制者

import java.net.URL;
import java.util.ResourceBundle;
import javafx.application.HostServices;
import javafx.event.ActionEvent;
import javafx.fxml.FXML;
import javafx.fxml.Initializable;
import javafx.scene.control.Label;

/**
 *
 * @author sedri
 */
public class FXMLDocumentController implements Initializable {

    HostServices hostServices;

    @FXML
    private Label label;

    @FXML
    private void handleButtonAction(ActionEvent event) {
        hostServices.showDocument("www.google.com");
    }

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

    public void setGetHostController(HostServices hostServices)
    {
        this.hostServices = hostServices;
    }

}

XML文件

<?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" prefWidth="320" xmlns:fx="http://javafx.com/fxml/1" fx:controller="javafxapplication7.FXMLDocumentController">
    <children>
        <Button layoutX="126" layoutY="90" text="Click Me!" onAction="#handleButtonAction" fx:id="button" />
        <Label layoutX="126" layoutY="120" minHeight="16" minWidth="69" fx:id="label" />
    </children>
</AnchorPane>

解决方法

有什么简单的方法可以将getHostServices（）。showDocument（）命令以某种方式放入toHomepage（）方法中，而不是执行几行代码，因此代码看起来应该很简洁？

package sample;

import javafx.application.HostServices;
import javafx.event.ActionEvent;
import javafx.fxml.FXML;
import javafx.scene.control.Button;

public class Controller {

   @FXML
   private Button facebookButton;

   @FXML
   void toHomepage(ActionEvent event) {

   }


}

如果我按下按钮，它应该直接将我链接到Facebook URL