Anish Sharma Anish Sharma - 7 months ago 47
Java Question

When Click on Hyperlinks in JavaFX ,a relevant URL should open in browser

I am developing an application in which I have some links added to the Listview and these links will keep on adding at runtime on some condition.So what I can't find is a way to how to open a url when clicking on a particular link.

This is the code for adding links into list view

if(counter==1)
{
Task task2 = new Task<Void>() {
@Override
public Void call() throws Exception {

Platform.runLater(new Runnable() {
public void run() {
link=new Hyperlink(val);
link.setStyle("-fx-border-style: none;");
items.add(link);
listview.setItems(items);



}
});


return null;

}
};
Thread th = new Thread(task2);
th.setDaemon(true);
th.start();
Thread.sleep(1000);


}


I know I need to use something like this to open a url in browser when click on link

getHostServices().showDocument(link.getText());


but I don't know how to listen/track the click event for different links

Answer

I made a very small example Application for you,

import java.util.ArrayList;
import java.util.List;
import javafx.application.Application;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Hyperlink;
import javafx.scene.control.ListView;
import javafx.scene.control.TextField;
import javafx.scene.layout.AnchorPane;
import javafx.scene.layout.HBox;
import javafx.scene.layout.VBox;
import javafx.stage.Stage;

public class ListList extends Application {

    final ListView listView = new ListView();
    @Override
    public void start(Stage primaryStage) {

        List<Hyperlink> links = new ArrayList<>();

        AnchorPane pane = new AnchorPane();
        VBox vBox = new VBox();
        final Hyperlink link = new Hyperlink("http://blog.professional-webworkx.de");
        Hyperlink link2= new Hyperlink("http://www.stackoverflow.com");

        links.add(link);
        links.add(link2);

        for(final Hyperlink hyperlink : links) {
            hyperlink.setOnAction(new EventHandler<ActionEvent>() {

                @Override
                public void handle(ActionEvent t) {
                    getHostServices().showDocument(hyperlink.getText());
                }
            });
        }

        listView.getItems().addAll(links);
        HBox hBox = new HBox();
        final TextField urlField = new TextField();
        Button b = new Button("Add Links");
        hBox.getChildren().addAll(b, urlField);

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

            @Override
            public void handle(ActionEvent t) {
                addLink(urlField.getText().trim());
                urlField.clear();
            }
        });
        vBox.getChildren().add(hBox);
        vBox.getChildren().add(listView);
        pane.getChildren().add(vBox);
        Scene scene = new Scene(pane, 800, 600);
        primaryStage.setTitle("Hello World!");
        primaryStage.setScene(scene);
        primaryStage.show();
    }

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

    private void addLink(final String url) {
        final Hyperlink link = new Hyperlink(url);
        link.setOnAction(new EventHandler<ActionEvent>() {

            @Override
            public void handle(ActionEvent t) {
                getHostServices().showDocument(link.getText());
                //openBrowser(link.getText());
            }

        });
        listView.getItems().add(link);
    }

    private void openBrowser(final String url) {
        getHostServices().showDocument(url);
    }
}

If you enter a new URL in the TextField and click on the Button, the new Link will be added to your LinkList and will be displayed on the ListView. Everytime you add a new Link, the .setOnAction() Method is set with the right URL to open.

Maybe you can use this as starting point for further developing your app.

Patrick