Question

我正在开发android应用程序，用户可以将视频上传到服务器。我是android开发的新手，所以我无法掌握上传视频的概念。我知道的是，我需要创建一个意图让用户选择视频，然后将其转换为uri文件。我打算将asynctask与httpurlconnection一起使用。谁能帮我处理asynctask？

Answer 1

我认为这对您有帮助 upload video

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

Android将视频分段上传到服务器

1 个答案: