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2020-06-27 10:49发布
乱世女痞
I want to upload video on specific server in Android.
Is is possible ?
How I can achieve this ?
Thanks.
Had the same issue some time ago. Here's a code.
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
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Had the same issue some time ago. Here's a code.