I am creating an application in JavaFx, In which I want to do that if any child stage is getting opened then it should be opened in center of parent stage.
I am trying to do this using mystage.centerOnScreen()
but it'll assign the child stage to center of screen, not the center of parent stage.
How can I assign the child stage to center of parent stage?
private void show(Stage parentStage) {
mystage.initOwner(parentStage);
mystage.initModality(Modality.WINDOW_MODAL);
mystage.centerOnScreen();
mystage.initStyle(StageStyle.UTILITY);
mystage.show();
}
You can use the parent stage's X/Y/width/height properties to do that. Rather than using Stage#centerOnScreen
, you could do the following:
public class CenterStage extends Application {
@Override
public void start(final Stage stage) throws Exception {
stage.setX(300);
stage.setWidth(800);
stage.setHeight(400);
stage.show();
final Stage childStage = new Stage();
childStage.setWidth(200);
childStage.setHeight(200);
childStage.setX(stage.getX() + stage.getWidth() / 2 - childStage.getWidth() / 2);
childStage.setY(stage.getY() + stage.getHeight() / 2 - childStage.getHeight() / 2);
childStage.show();
}
public static void main(String[] args) {
Application.launch(args);
}
}
When you don't determine a size for the childStage, you have to listen for width and height changes as width and height is still NaN when onShown is called.
final double midX = (parentStage.getX() + parentStage.getWidth()) / 2;
final double midY = (parentStage.getY() + parentStage.getHeight()) / 2;
xResized = false;
yResized = false;
newStage.widthProperty().addListener((observable, oldValue, newValue) -> {
if (!xResized && newValue.intValue() > 1) {
newStage.setX(midX - newValue.intValue() / 2);
xResized = true;
}
});
newStage.heightProperty().addListener((observable, oldValue, newValue) -> {
if (!yResized && newValue.intValue() > 1) {
newStage.setY(midY - newValue.intValue() / 2);
yResized = true;
}
});
newStage.show();