我想允许用户使用按钮进入下一个窗口 . 然后,此按钮显示图像 . 但是,当我尝试运行程序时,我会抛出一个异常 . 这是我的代码:
package matchingcards;
import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.image.Image;
import javafx.scene.image.ImageView;
import javafx.scene.layout.GridPane;
import javafx.stage.Stage;
public class MatchingCards extends Application {
Stage window;
Scene start, game;
public static void main(String[] args) {
launch(args);
}
@Override
public void start(Stage primaryStage) throws Exception {
window = primaryStage;
// When user clicks start, program enters game
Button btn = new Button("Start Game");
btn.setOnAction(e -> window.setScene(game));
// Create the display for start button
GridPane pane1 = new GridPane();
pane1.getChildren().addAll(pane1, btn);
start = new Scene(pane1, 200, 200);
// Create display for game
GridPane pane2 = new GridPane();
Image back = new
Image("https://i.pinimg.com/736x/c1/59/b4/" +
"c159b4738dae9c9d8d6417228024de8d--" +
"playing-card-design-card-card.jpg", 300, 200, false, false);
pane2.getChildren().addAll(new ImageView(back));
Scene game = new Scene(pane2, 500, 500);
window.setScene(start);
window.setTitle("Matching Cards");
window.show();
}
}
1 回答
您遇到异常,因为您在其中添加了pane1
我已经纠正了你的代码: