音乐播放脚本12个错误

-2

这是一部分，我认为我有就麻烦了：音乐播放脚本12个错误

public class AudioPlayer { 
    //play the music 
    AudioPlayer AP = new AudioPlayer([Optional Replay = true]); 
    AP.playSong(signlink.findcachedir() + "Music/2.wav"); 
    //stop the music 
    AP.Stop(); 
}

下面是我得到的错误：

AudioPlayer.java:3: error: ')' expected AudioPlayer AP = new AudioPlayer(optional Replay = true); ^AudioPlayer.java:3: error: illegal start of type AudioPlayer AP = new AudioPlayer(optional Replay = true); ^AudioPlayer.java:3: error: expected AudioPlayer AP = new AudioPlayer(optional Replay = true); ^AudioPlayer.java:3: error: ';' expected AudioPlayer AP = new AudioPlayer(optional Replay = true); ^AudioPlayer.java:4: error: expected AP.playSong(signlink.findcachedir() + "Music/2.wav"); ^AudioPlayer.java:4: error: expected AP.playSong(signlink.findcachedir() + "Music/2.wav"); ^AudioPlayer.java:4: error: ';' expected AP.playSong(signlink.findcachedir() + "Music/2.wav"); ^AudioPlayer.java:4: error: illegal start of type AP.playSong(signlink.findcachedir() + "Music/2.wav"); ^AudioPlayer.java:4: error: expected AP.playSong(signlink.findcachedir() + "Music/2.wav"); ^AudioPlayer.java:4: error: ';' expected AP.playSong(signlink.findcachedir() + "Music/2.wav"); ^AudioPlayer.java:6: error: expected AP.Stop(); ^AudioPlayer.java:6: error: reached end of file while parsing AP.Stop(); ^12 errors Press any key to continue . . .`

正在发生什么的任何想法？什么是导致错误，我该如何解决它？

来源

2015-01-05 jake

AudioPlayer类构造函数采用哪些参数？ –

你在那里有没有有效的语法。你只是......不能在Java中这样做。你试图接受一个布尔到该构造函数？ – Makoto

类，接口或枚举 – jake

-1

尝试用这个来更换此行

AudioPlayer AP = new AudioPlayer([Optional Replay = true]);

：

boolean optionalReplay = true; 
AudioPlayer AP = new AudioPlayer(optionalReplay);

程序不理解这样一句话：[Optional Replay = true]。我给你的是更适合Java的语法。

仅供参考，我假设AudioPlayer构造函数需要一个布尔参数。但是我可能猜测你写代码的方式是错误的。

来源

2015-01-05 00:53:51

谢谢，修复了大量的错误我现在降到8错误 – jake

酷，请注册并接受。你的8个错误是什么？ –

仔细看看代码。这样做并不能解决问题。 – Makoto

音乐播放脚本12个错误

回答

相关问题