Question

我已经捕获了一个视频，我想通过webservice调用上传到服务器。我已经尝试过使用volley库，即我们如何上传图像我已经尝试了同样的视频，但它没有上传。请给我一些好的库用于在Android上传视频。

谢谢， Rojesh

Answer 1

使用以下功能将视频上传到服务器...

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://your_website.com/upload_audio_test/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);
        }
      }

PHP网络服务。

<?php
// Where the file is going to be placed
$target_path = "uploads/";

/* Add the original filename to our target path.
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']).
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
    echo "filename: " .  basename( $_FILES['uploadedfile']['name']);
    echo "target_path: " .$target_path;
}
?>

如何通过Webservice将Android中的视频上传到服务器？

1 个答案: