这最初是由Martin Dow
import java.io.*;
import javax.sound.sampled.*;
class AudioFileProcessor {
public static void main(String[] args) {
copyAudio("/tmp/uke.wav", "/tmp/uke-shortened.wav", 2, 1);
}
public static void copyAudio(String sourceFileName, String destinationFileName, int startSecond, int secondsToCopy) {
AudioInputStream inputStream = null;
AudioInputStream shortenedStream = null;
try {
File file = new File(sourceFileName);
AudioFileFormat fileFormat = AudioSystem.getAudioFileFormat(file);
AudioFormat format = fileFormat.getFormat();
inputStream = AudioSystem.getAudioInputStream(file);
int bytesPerSecond = format.getFrameSize() * (int)format.getFrameRate();
inputStream.skip(startSecond * bytesPerSecond);
long framesOfAudioToCopy = secondsToCopy * (int)format.getFrameRate();
shortenedStream = new AudioInputStream(inputStream, format, framesOfAudioToCopy);
File destinationFile = new File(destinationFileName);
AudioSystem.write(shortenedStream, fileFormat.getType(), destinationFile);
} catch (Exception e) {
println(e);
} finally {
if (inputStream != null) try { inputStream.close(); } catch (Exception e) { println(e); }
if (shortenedStream != null) try { shortenedStream.close(); } catch (Exception e) { println(e); }
}
}
}
最初回答HERE
仅供参考,大多数回答。 wav文件是44.1KHz,意味着每个样本持续超过2000ns。你不会得到毫微秒的精度 –
你已经做了什么来解决这个问题?你在寻找现有解决方案时做了哪些研究? – Asaf
@阿萨夫可能你没有读过这个问题。你只能阅读标题! –