我正在开发一个 android 应用程序,让用户录制 30 秒长的视频,然后我需要将此视频文件保存在服务器上。我正在使用一个接受 byteArray 格式的视频文件的网络服务。所以,我不知道如何转换和压缩 .mp4 文件,然后通过网络将其发送到 Web 服务。请帮忙...
问问题
2787 次
1 回答
1
尝试
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
DataInputStream inputStream = null;
String pathOfYourFile = "/sdcard/videoName.3gp";
String urlServer = "http://....../uploadvideo.php";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
try
{
FileInputStream fileInputStream = new FileInputStream(new File(pathOfYourFile) );
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Enable POST method
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary");
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + pathOfYourFile );
outputStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Read file
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
outputStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outputStream.writeBytes(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
Log.d("ServerCode",""+serverResponseCode);
Log.d("serverResponseMessage",""+serverResponseMessage);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
ex.printStackTrace();
}
}
于 2013-07-16T19:43:49.480 回答