Question

我的参考资料部分有大约20个小的声音效果文件...我想有一个方法根据我传递的参数播放特定的声音文件...假设我的声音文件被命名sound01.wav，sound02.wav等。

public static void PlayMySound(string soundFile)
{
   SoundPlayer snd = new SoundPlayer(Properties.Resources.XXX);
   snd.Play()
}

PlayMySound(sound01);
PlayMySound(sound02);
etc.

在上面的代码中，我希望XXX成为字符串soundFile

我试图避免这样的事情......

public static void PlayMySound(string soundFile)
{
   if (soundFile == "sound01") { SoundPlayer snd = new SoundPlayer(Properties.Resources.sound01); }
   if (soundFile == "sound02") { SoundPlayer snd = new SoundPlayer(Properties.Resources.sound02); }
   if (soundFile == "sound03") { SoundPlayer snd = new SoundPlayer(Properties.Resources.sound03); }
   if (soundFile == "sound04") { SoundPlayer snd = new SoundPlayer(Properties.Resources.sound04); }
   if (soundFile == "sound05") { SoundPlayer snd = new SoundPlayer(Properties.Resources.sound05); }
   etc.etc.etc.
}

Answer 1

好的，我在发布此消息后不久就回答了我......

public static void PlayMySound(Stream soundFile)
{
   SoundPlayer snd = new SoundPlayer(soundFile);
   snd.Play();
}

// Then I can just call it like this...
PlayMySound(Properties.Resources.sound01);
PlayMySound(Properties.Resources.sound02);

使用参数指定参考资源中的哪个声音文件

1 个答案: