Question

相当新的Java并尝试自学一点关于JavaFX。我正在尝试创建一个单击视频文件时运行的简单JavaFX视频/媒体播放器。我想将实际播放器创建为一个单独的类，它接受视频文件位置作为字符串参数。

当我在下面运行时，

public class BLPlayer{
    public static void main(String[] args) {
        if(args.length > 0){
            VideoPlayer vp = new VideoPlayer(args);
        }else{
            //showGUI();
        }
    }
}


public class VideoPlayer extends Application {
    String path;
    MediaPlayer player;
    Scene scene;
    MediaView view;
    Group root;
    Media media;

    VideoPlayer(String[] args){
        path = args[0];
        path = path.replace("\\", "/"); 
        launch(args);
    }

    @Override
    public void start(final Stage stage) throws Exception {

        File f = new File(path);

        root = new Group();
        media = new Media(f.toURI().toString());
        player = new MediaPlayer(media);
        view = new MediaView(player);
        root.getChildren().add(view);
        scene = new Scene(root, 400, 400, Color.BLACK);
        stage.setScene(scene);
        stage.show();
        player.play();
    }

}

我收到错误：

Exception in Application constructor
Exception in thread "main" java.lang.RuntimeException: Unable to construct Application instance: class player.VideoPlayer
at com.sun.javafx.application.LauncherImpl.launchApplication1(LauncherImpl.java:884)
at com.sun.javafx.application.LauncherImpl.access$000(LauncherImpl.java:56)
at com.sun.javafx.application.LauncherImpl$1.run(LauncherImpl.java:158)
at java.lang.Thread.run(Thread.java:745)
Caused by: java.lang.NoSuchMethodException: player.VideoPlayer.<init>()
at java.lang.Class.getConstructor0(Class.java:2971)
at java.lang.Class.getConstructor(Class.java:1812)
at com.sun.javafx.application.LauncherImpl$7.run(LauncherImpl.java:790)
at com.sun.javafx.application.PlatformImpl$7.run(PlatformImpl.java:335)
at com.sun.javafx.application.PlatformImpl$6$1.run(PlatformImpl.java:301)
at com.sun.javafx.application.PlatformImpl$6$1.run(PlatformImpl.java:298)
at java.security.AccessController.doPrivileged(Native Method)
at com.sun.javafx.application.PlatformImpl$6.run(PlatformImpl.java:298)
at com.sun.glass.ui.InvokeLaterDispatcher$Future.run(InvokeLaterDispatcher.java:95)
at com.sun.glass.ui.win.WinApplication._runLoop(Native Method)
at com.sun.glass.ui.win.WinApplication.access$300(WinApplication.java:39)
at com.sun.glass.ui.win.WinApplication$4$1.run(WinApplication.java:112)

我不知道为什么它不起作用。我很感激任何提示或建议！谢谢你的帮助。

Answer 1

发生错误是因为您需要使用launch()启动JavaFX应用程序。有关详情go through this solution。

如果您确实需要将参数发送到VideoPlayer类。您可以使用getParameters()类的Application来获取参数。

public class VideoPlayer extends Application {
    String path;
    MediaPlayer player;
    Scene scene;
    MediaView view;
    Group root;
    Media media;

    @Override
    public void start(final Stage stage) throws Exception {
        Parameters params = getParameters();
        final List<String> parameters = params.getRaw();
        path = !parameters.isEmpty() ? parameters.get(0) : "";
        path = path.replace("\\", "/"); 
        root = new Group();
        File f = new File(path);
        root = new Group();
        media = new Media(f.toURI().toString());
        player = new MediaPlayer(media);
        view = new MediaView(player);
        root.getChildren().add(view);
        scene = new Scene(root, 400, 400, Color.BLACK);
        stage.setScene(scene);
        stage.show();
        player.play();
    }
}

从另一个类

启动JavaFX应用程序

public class BLPlayer {
    public static void main(String[] args) {
        if(args.length > 0){
            Application.launch(VideoPlayer.class, args);
        }
    }
}

创建JavaFX Video Player类的实例

1 个答案: