我正在尝试基于视频的应用程序。我使用下面的方法上传视频文件,当我发送大尺寸的视频文件时,它有更多的时间上传。所以我需要压缩数据,并且视频文件大小应该小于30 MB。我的英文不好原谅我。谁能给我解决我的问题的方法呢?
private void doFileUpload(){
HttpURLConnection conn = null;
DataOutputStream dos = null;
DataInputStream inStream = null;
String lineEnd = "rn";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
String responseFromServer = "";
String urlString = "http://mywebsite.com/upload_audio.php";
try
{
//------------------ CLIENT REQUEST
FileInputStream fileInputStream = new FileInputStream(new File(selectedPath) );
// open a URL connection to the Servlet
URL url = new URL(urlString);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
// Allow Inputs
conn.setDoInput(true);
// Allow Outputs
conn.setDoOutput(true);
// Don't use a cached copy.
conn.setUseCaches(false);
// Use a post method.
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
dos = new DataOutputStream( conn.getOutputStream() );
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name="uploadedfile";filename="" + selectedPath + """ + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// close streams
Log.e("Debug","File is written");
fileInputStream.close();
dos.flush();
dos.close();
}
catch (MalformedURLException ex)
{
Log.e("Debug", "error: " + ex.getMessage(), ex);
}
catch (IOException ioe)
{
Log.e("Debug", "error: " + ioe.getMessage(), ioe);
}
//------------------ read the SERVER RESPONSE
try {
inStream = new DataInputStream ( conn.getInputStream() );
String str;
while (( str = inStream.readLine()) != null)
{
Log.e("Debug","Server Response "+str);
}
inStream.close();
}
catch (IOException ioex){
Log.e("Debug", "error: " + ioex.getMessage(), ioex);
}
}
答案 0 :(得分:0)
您可以尝试HttpClient jar下载最新的HttpClient jar,将其添加到您的项目中,然后使用以下方法上传视频:
private void uploadVideo(String videoPath) throws ParseException, IOException {
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(YOUR_URL);
FileBody filebodyVideo = new FileBody(new File(videoPath));
StringBody title = new StringBody("Filename: " + videoPath);
StringBody description = new StringBody("This is a video of the agent");
StringBody code = new StringBody(realtorCodeStr);
MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("videoFile", filebodyVideo);
reqEntity.addPart("title", title);
reqEntity.addPart("description", description);
reqEntity.addPart("code", code);
httppost.setEntity(reqEntity);
// DEBUG
System.out.println( "executing request " + httppost.getRequestLine( ) );
HttpResponse response = httpclient.execute( httppost );
HttpEntity resEntity = response.getEntity( );
// DEBUG
System.out.println( response.getStatusLine( ) );
if (resEntity != null) {
System.out.println( EntityUtils.toString( resEntity ) );
} // end if
if (resEntity != null) {
resEntity.consumeContent( );
} // end if
httpclient.getConnectionManager( ).shutdown( );
} // end of uploadVideo( )