Set custom control's Button "onAction" method in the FXML file?

Question

I created a new custom control containing labels, text field and a button. How can I set custom control's Button "onAction" method in the FXML file?

Example code:

<BorderPane fx:controller="fxmlexample.MyController" xmlns:fx="http://javafx.com/fxml"> <top> <MyCustomComponent onButtonAction="#myCustomButtonAction"> </MyCustomComponent> </top>

You need to expose a method from your custom control which fire's the button and then use it in FXML. — ItachiUchiha
– ItachiUchiha, Commented Aug 12, 2016 at 11:39

Community · Accepted Answer · 2017-05-23 12:33:03Z

Ok, I found a solution. I did it exactly as Sergey Grinev suggested in this thread - JavaFX 2.0 - create action handler for custom component in FXML. The only change I made was using my button's field name instead of "this" in Class constructor.

fabian · Accepted Answer · 2016-08-12 12:13:41Z

onButtonAction="#myCustomButtonAction" can be used as long as MyCustomComponent contains a setter for the event handler named setOnButtonAction.

class MyCustomComponent { private Button button; public void setOnButtonAction(EventHandler<ActionEvent> handler) { this.button.setOnAction(handler); } public EventHandler<ActionEvent> getOnButtonAction() { return this.button.getOnAction(); } ... }

Unfortunately, this did not work. But I found a solution (in my answer).
@notmyf4ulty sorry, forgot that the getter is required for the FXMLLoader to get handler type. Should work now...

Collectives™ on Stack Overflow

Set custom control's Button "onAction" method in the FXML file?

2 Answers 2

Comments

2 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Linked

Related