我现在写个应用程序来播放声音文件 ，播放单个声音文件没有问题 ，
但我连续播放多个声音文件 却有问题 ，我不想用 Thread.sleep(300) 这样实现文件与文件的间隔
我将 多个文件的字节流合并成一个 在播放去只播放第一个文件 不知道哪里有问题
希望大家指导，帮助
代码：
public class Sound {
private AudioStream as = null;
private byte[] bs = null;
public Sound() {
}
/**
* 播放单个文件
*/
public synchronized void play(InputStream ins) {
try {
as = new AudioStream(ins);
} catch (IOException e) {
e.printStackTrace();
}
play(as);
}
/**
* 播放
*
* @param as
*/
public synchronized void play(AudioStream as) {
AudioPlayer.player.start(as);
try {
AudioPlayer.player.join(5000L);
} catch (InterruptedException e) {
e.printStackTrace();
}
AudioPlayer.player.stop(as);
}
/**
* 播放单个文件
*/
public synchronized void play(InputStream[] ins) {
as = getAllAudioStream(ins);
play(as);
}
/**
* 将多个流合并
*
* @param ins
* @return
*/
private AudioStream getAllAudioStream(InputStream[] ins) {
try {
if (ins == null || ins.length == 0) {
return null;
}
byte[] bytes = new byte[1024100];
bs = new byte[0];
for (int i = 0, count = ins.length; i < count; i++) {
InputStream in = ins[i];
int length = 0;
length = in.read(bytes);
byte[] tmp = new byte[bs.length + length];
int j = 0;
for (; j < bs.length; j++) {
tmp[j] = bs[j];
}
for (int k = 0; k < length; k++) {
tmp[j++] = bytes[k];
}
bs = new byte[tmp.length];
for (int n = 0; n < tmp.length; n++) {
bs[n] = tmp[n];
}
}
as = new AudioStream(new ByteArrayInputStream(bs));
} catch (IOException e) {
e.printStackTrace();
}
return as;
}
}
不建议合并流
推荐楼主递归调用单个播放的方法