从ServletInputStream获取wav文件

发布于 2024-11-30 22:53:59 字数 2560 浏览 1 评论 0原文

我们当前的项目需要我们发送一个音频文件到服务器,然后使用该音频文件进行进一步的计算。

使用 Java 声音 api,我能够捕获录音并将其保存为系统中的 wav 文件。然后,为了将音频 wav 传递到服务器,我使用 Apache Commons HttpClient 向服务器发送请求。 (我正在使用 apache 提供的 InputstreamEntity 并将数据作为块发送)。

当我尝试在服务器上重新创建/检索 wav 文件时出现问题。我知道我必须使用 AudioSystem.write API 来创建 wav 文件(与我的系统上所做的完全相同)。然而,我观察到,虽然文件已创建,但它无法播放(我正在使用 vlc 媒体播放器来测试它,仅供参考)。我在 Google 中搜索了示例代码并尝试实现它,但是一旦创建文件就无法播放它。

示例代码片段表明了我尝试过的方法:

//******************************************************************

try {
InputStream is = request.getInputStream();
FileOutputStream fs = new FileOutputStream("output123.wav");
byte[] tempbuffer = new byte[4096];
int bytesRead;
while((bytesRead=is.read(tempbuffer))!=-1)
{
fs.write(tempbuffer, 0,bytesRead);
}
is.close();
fs.close();
AudioInputStream inputStream =AudioSystem.getAudioInputStream(newFile("output123.wav"));
int numofbytes = inputStream.available();           
byte[] buffer = new byte[numofbytes];
inputStream.read(buffer);
int bytesWritten = AudioSystem.write(inputStream, AudioFileFormat.Type.WAVE,new File("outputtest.wav"));
System.out.println("written"+bytesWritten);

方法 2

InputStream is = request.getInputStream(); 
System.out.println("inputStream obtained : "+is.toString());
ByteArrayInputStream bais = null;
byte[] audioBuffer = IOUtils.toByteArray(is);
System.out.println(" is audioBuffer empty? : length =  ? "+audioBuffer.length);
try {
AudioFileFormat ai = AudioSystem.getAudioFileFormat(is);
System.out.println("ai bytelength ? "+ai.getByteLength());
System.out.println("ai frame length = "+ai.getFrameLength());
Set<Map.Entry<String,Object>> audioProperties = ai.getFormat().properties().entrySet();
System.out.println("entry set is empty ? "+audioProperties.isEmpty());          
for(Map.Entry me : audioProperties){
System.out.println("key = "+me.getKey());
System.out.println("value ="+me.getValue());}       
bais = new ByteArrayInputStream(audioBuffer);
AudioInputStream ais = new AudioInputStream(bais, new AudioFormat(8000,8,2,true,true), 2);
AudioSystem.write(ais, AudioFileFormat.Type.WAVE,new File("testtest.wav"));

//*************************************************************************************

audioFormat 属性全部结果为 null。这些空值是否会带来问题?因此,在服务器上创建波形文件时,我尝试再次手动设置属性。但即使如此,wav 文件也无法播放。

我还尝试了本网站上已经提到的相当多的方法,但不知何故它们不起作用。我确信我错过了一些东西,但我无法查明确切的问题。

如果你们能指出如何从 ServletInputStream 转换为获取 wav,那将非常有帮助。

PS (1) 我知道代码很破旧,因为我已经处于试错状态很长一段时间了。但如果需要的话,我会提供有关这些方法的更多细节。 2)为我的笨拙道歉,这恰好是我的第一篇文章..)

Our current project requires us to send an audio file to the server and then use the audio file for further computation.

Using the Java sound api, I was able to capture the recording and save it as a wav file in my system. Then in order to pass the audio wav to the server, I am using Apache Commons HttpClient to post a request to the server. (I am using InputstreamEntity provided by apache and sending the data as a chunk).

The problem appears when i am trying to recreate/retrieve the wav file on the server. I understand that I would have to use the AudioSystem.write API to create the wav file (exactly as what was done on my system). However what I observe is that althought the file gets created , it does not play (I am using vlc media player to test it FYI). I have searched in Google for sample codes and have tried to implement it, but is unable to play it once the file gets created.

The sample code snippets indicates the approaches i have tried:

//******************************************************************

try {
InputStream is = request.getInputStream();
FileOutputStream fs = new FileOutputStream("output123.wav");
byte[] tempbuffer = new byte[4096];
int bytesRead;
while((bytesRead=is.read(tempbuffer))!=-1)
{
fs.write(tempbuffer, 0,bytesRead);
}
is.close();
fs.close();
AudioInputStream inputStream =AudioSystem.getAudioInputStream(newFile("output123.wav"));
int numofbytes = inputStream.available();           
byte[] buffer = new byte[numofbytes];
inputStream.read(buffer);
int bytesWritten = AudioSystem.write(inputStream, AudioFileFormat.Type.WAVE,new File("outputtest.wav"));
System.out.println("written"+bytesWritten);

Approach 2

InputStream is = request.getInputStream(); 
System.out.println("inputStream obtained : "+is.toString());
ByteArrayInputStream bais = null;
byte[] audioBuffer = IOUtils.toByteArray(is);
System.out.println(" is audioBuffer empty? : length =  ? "+audioBuffer.length);
try {
AudioFileFormat ai = AudioSystem.getAudioFileFormat(is);
System.out.println("ai bytelength ? "+ai.getByteLength());
System.out.println("ai frame length = "+ai.getFrameLength());
Set<Map.Entry<String,Object>> audioProperties = ai.getFormat().properties().entrySet();
System.out.println("entry set is empty ? "+audioProperties.isEmpty());          
for(Map.Entry me : audioProperties){
System.out.println("key = "+me.getKey());
System.out.println("value ="+me.getValue());}       
bais = new ByteArrayInputStream(audioBuffer);
AudioInputStream ais = new AudioInputStream(bais, new AudioFormat(8000,8,2,true,true), 2);
AudioSystem.write(ais, AudioFileFormat.Type.WAVE,new File("testtest.wav"));

//*************************************************************************************

The audioFormat properties all turned out to be null. Are these null values giving the problem? So while creating the wave file on the server, I tried to set the properties manually once again. But even then the wav file would not play.

I have also tried quite a few approaches already mentioned on this site, but somehow they aren't working. I am sure i am missing something, but I am unable to pinpoint the exact problem.

Would be really helpful, if you guys can point out how to go about the conversion from ServletInputStream to getting a wav.

P.S (1) I know the code is shabby, because i have been under a trial and error situation for quite some time now. But I will give more details on the approaches if needed.
2) Apologise for the clumsiness, this happens to be my first post.. )

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评论(3

烧了回忆取暖 2024-12-07 22:53:59

这不是复制流的方式(来自方法 1)。您有正确的代码来复制上面的流:

int numofbytes = inputStream.available();           
byte[] buffer = new byte[numofbytes];
inputStream.read(buffer);

this is not how you copy a stream (from Approach 1). you have the correct code to copy a stream just above this.:

int numofbytes = inputStream.available();           
byte[] buffer = new byte[numofbytes];
inputStream.read(buffer);
我的鱼塘能养鲲 2024-12-07 22:53:59

如果您的服务器想要做的只是获取数据并将其写入文件,那么您不需要需要使用任何音频 API:只需将数据视为字节流即可。

因此,方法 1 中提到 AudioInputStream 之前的部分就足够了。

If all your server wants to do is get the data and write it to a file, then you do not need to use any of the audio API: simply treat the data as a stream of bytes.

So the part of approach 1 that is before any mention of AudioInputStream should be sufficient.

舟遥客 2024-12-07 22:53:59

虽然选择的方法可能不是完美的解决方案,但由于时间限制,我采用了更简单的方法。使用 java.util.zip 我只是将其压缩并将其发送到服务器,然后在文件解压缩的地方编写一个层。然后我删除了zip文件。似乎是一个不成熟的解决方案(因为最初的挑战是发送音频文件)。现在我需要压缩文件,但文件传输会相对更快。谢谢你们的帮助。

Although the approach chosen might not be the perfect solution, due to time constraints, I adopted a simpler approach. Using java.util.zip i simply zipped it up and sent it over to the server and then wrote a layer wherin the file gets unzipped . then i deleted the zip files. Seems like an immature solution (bcos the original challenge was to send the audio file). now i am incurring an overhead of zipping the files, but the file transfer would hapeen relatively faster. Thanks for your help guys.

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