Upload progress bar in android?

Ich verwende den folgenden code, um das Video hochzuladen, um php-server, es war in Ordnung arbeiten, aber ich brauche, um den Fortschritt zu zeigen-bar, während die Datei hochgeladen wird, muss ich aus der Synchronisation von Datei-upload und die Fortschrittsleiste erhöhen.Jemand deuten einige Idee?

      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;
      }
      int serverResponseCode=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("uploadedfile", fileName);

      //conn.setFixedLengthStreamingMode(1024);
       //conn.setChunkedStreamingMode(1);
       dos = new DataOutputStream(conn.getOutputStream());

       dos.writeBytes(twoHyphens + boundary + lineEnd);
       dos.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";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);
       bufferSize=(int)sourceFile.length();


       System.out.println("BytesAvail"+bytesAvailable);
       System.out.println("maxBufferSize"+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;  //like 200 (Ok) 

Dank .

Überprüfen Sie die folgende Antwort, vielleicht hilft es: [stackoverflow.com/questions/15572747/... [1]: stackoverflow.com/a/19295719/463846
Wie können Sie zusätzliche Daten, wie ein string, mit dieser multipart-Daten?

InformationsquelleAutor Karthi | 2011-02-24

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