Question

我想在Android上的特定服务器上传视频。

有可能吗？

我怎么能做到这一点？

感谢。

Answer 1

前段时间有同样的问题。这是一个代码。

public static int upLoad2Server(String sourceFileUri)
{
 String upLoadServerUri = "your remote server link";
 // String [] string = sourceFileUri;
 String fileName = sourceFileUri;

 HttpURLConnection conn = null;
 DataOutputStream dos = null;
 DataInputStream inStream = null;
 String lineEnd = "\r\n";
 String twoHyphens = "--";
 String boundary = "*****";
 int bytesRead, bytesAvailable, bufferSize;
  byte[] buffer;
 int maxBufferSize = 1 * 1024 * 1024;
 String responseFromServer = "";

 File sourceFile = new File(sourceFileUri);
 if (!sourceFile.isFile()) {
 Log.e("Huzza", "Source File Does not exist");
 return 0;
  }
try { // open a URL connection to the Servlet
 FileInputStream fileInputStream = new FileInputStream(sourceFile);
 URL url = new URL(upLoadServerUri);
 conn = (HttpURLConnection) url.openConnection(); // Open a HTTP  connection to  the URL 
 conn.setDoInput(true); // Allow Inputs
 conn.setDoOutput(true); // Allow Outputs
 conn.setUseCaches(false); // Don't use a Cached Copy
 conn.setRequestMethod("POST");
 conn.setRequestProperty("Connection", "Keep-Alive");
 conn.setRequestProperty("ENCTYPE", "multipart/form-data");
 conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
 conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());

dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""+ fileName + "\"" + lineEnd);
 dos.writeBytes(lineEnd);

 bytesAvailable = fileInputStream.available(); // create a buffer of  maximum size
 Log.i("Huzza", "Initial .available : " + bytesAvailable);

 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);

 // Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();

Log.i("Upload file to server", "HTTP Response is : " + serverResponseMessage + ": " + serverResponseCode);
// close streams
Log.i("Upload file to server", fileName + " File is written");
 fileInputStream.close();
dos.flush();
dos.close();
 } catch (MalformedURLException ex) {
 ex.printStackTrace();
Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
} catch (Exception e) {
e.printStackTrace();
 }
//this block will give the response of upload link
  try {
  BufferedReader rd = new BufferedReader(new InputStreamReader(conn
    .getInputStream()));
  String line;
  while ((line = rd.readLine()) != null) {
   Log.i("Huzza", "RES Message: " + line);
  }
  rd.close();
 } catch (IOException ioex) {
  Log.e("Huzza", "error: " + ioex.getMessage(), ioex);
 }
 return serverResponseCode;  // like 200 (Ok)

} // end upLoad2Server

在Android上传视频

1 个答案: