如何检测舞台外的鼠标点击事件并关闭它

问题描述

我有一个充当弹出窗口的舞台，如果用户在该舞台外点击，我想关闭舞台，如何实现，舞台充当弹出窗口的代码如下：

I have a stage that acts as popup,I want to close the stage if the user clicks outside that stage,how to achieve that,My code for the stage acting as popup is as follows:

    Stage stage = new Stage();
    Parent root = FXMLLoader.load(getClass().getResource("colorchange.fxml"));
    Scene sce = new Scene(root,400,400);
    stage.initModality(Modality.APPLICATION_MODAL);
    stage.initOwner(color.getScene().getWindow());
    stage.setScene(sce);
    stage.showAndWait();

推荐答案

使用阶段注册一个监听器.focusedProperty（）并调用 stage.hide（）如果焦点属性更改为 false 。

Register a listener with stage.focusedProperty() and call stage.hide() if the focused property changes to false.

SSCCE：

import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.layout.StackPane;
import javafx.stage.Modality;
import javafx.stage.Stage;

public class CloseWindowOnClickOutside extends Application {

    @Override
    public void start(Stage primaryStage) {
        Button showPopup = new Button("Show popup");
        showPopup.setOnAction(e -> {
            Stage popup = new Stage();
            Scene scene = new Scene(new Label("Popup"), 120, 40);
            popup.setScene(scene);
            popup.initModality(Modality.APPLICATION_MODAL);
            popup.initOwner(primaryStage);
            popup.focusedProperty().addListener((obs, wasFocused, isNowFocused) -> {
                if (! isNowFocused) {
                    popup.hide();
                }
            });
            popup.show();
        });

        StackPane root = new StackPane(showPopup);
        Scene scene = new Scene(root, 350, 120);
        primaryStage.setScene(scene);
        primaryStage.show();
    }

    public static void main(String[] args) {
        launch(args);
    }
}

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