我正在使用javaFX在netbeans中编写程序 视图中有几个按钮,有一些坏按钮(比如炸弹是扫雷),我试图在按下坏按钮时冻结程序,但我找不到怎么做
谢谢!
答案 0 :(得分:0)
您的问题有各种解决方案。其中2个只是忽略动作事件或禁用这样的按钮:
public class ButtonAction extends Application {
final BooleanProperty buttonActionProperty = new SimpleBooleanProperty();
public static void main(String[] args) {
Application.launch(args);
}
@Override
public void start(Stage primaryStage) {
FlowPane root = new FlowPane();
CheckBox checkBox = new CheckBox( "Enabled");
checkBox.setSelected(true);
// solution 1: check if action is allowed and process it or not
buttonActionProperty.bind( checkBox.selectedProperty());
Button button = new Button( "Click Me");
button.setOnAction(e -> {
if( buttonActionProperty.get()) {
System.out.println( "Allowed, processing action");
} else {
System.out.println( "Not allowed, no action");
}
});
// solution 2: remove comments to activate the code
// button.disableProperty().bind(buttonActionProperty.not());
root.getChildren().addAll(checkBox, button);
primaryStage.setScene(new Scene(root, 600, 200));
primaryStage.show();
}
}
答案 1 :(得分:0)
添加一个ROOT类型的事件过滤器,它会消耗所有类型的事件(鼠标,键盘等)
btnThatHasHiddenMine.setOnAction(( ActionEvent event ) ->
{
System.out.println("Ohh no! You just stepped over the mine!");
getGameboardPane().addEventFilter( EventType.ROOT, Event::consume );
});
仅将过滤器添加到GameboardPane,因为我们不想冻结应用的其他部分。