Download file via HTTP Get in java

Question

I wrote a boot servlet to return a file based on the messageID parameter. The following is the doGet method.

@Override protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { // This messageID would be used to get the correct file eventually long messageID = Long.parseLong(request.getParameter("messageID")); String fileName = "C:\\Users\\Soto\\Desktop\\new_audio1.amr"; File returnFile = new File(fileName); ServletOutputStream out = response.getOutputStream(); ServletContext context = getServletConfig().getServletContext(); String mimetype = context.getMimeType("C:\\Users\\Soto\\Desktop\\new_audio1.amr"); response.setContentType((mimetype != null) ? mimetype : "application/octet-stream"); response.setContentLength((int)returnFile.length()); response.setHeader("Content-Disposition", "attachment; filename=\"" + "new_audio.amr" + "\""); FileInputStream in = new FileInputStream(returnFile); byte[] buffer = new byte[4096]; int length; while((length = in.read(buffer)) > 0) { out.write(buffer, 0, length); } in.close(); out.flush(); } 

Then I wrote code to extract the file.

 String url = "http://localhost:8080/AudioFileUpload/DownloadServlet"; String charset = "UTF-8"; // The id of the audio message requested String messageID = "1"; //URLConnection connection = null; try { String query = String.format("messageID=%s", URLEncoder.encode(messageID, charset)); //URLConnection connection; //URL u = new URL(url + "?" + query); //connection = u.openConnection(); //InputStream in = connection.getInputStream(); HttpClient httpClient = new DefaultHttpClient(); httpClient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1); HttpGet httpGet = new HttpGet(url + "?" + query); HttpResponse response = httpClient.execute(httpGet); System.out.println(response.getStatusLine()); InputStream in = response.getEntity().getContent(); FileOutputStream fos = new FileOutputStream(new File("C:\\Users\\Soto\\Desktop\\new_audio2.amr")); byte[] buffer = new byte[4096]; int length; while((length = in.read(buffer)) > 0) { fos.write(buffer, 0, length); } //connection = new URL(url + "?" + query).openConnection(); //connection.setRequestProperty("Accept-Charset", charset); //InputStream response = connection.getInputStream(); } catch (MalformedURLException e) { // TODO Auto-generated catch block e.printStackTrace(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } 

Now this code is working fine. I can upload an audio file and it works correctly. I want to know how, if possible, to get the file name as it is downloaded, instead of giving it my name. Also, is it possible to get a file without reading from a stream (maybe some kind of library that does this for you)? I kind of want to hide dirty things.

thanks

Answer 1

When you attach the file will be properly filed with the name provided. When inline, browsers seem to ignore the file name and usually provide the servlet name in the URL as the default name when saving inline content.

You can try matching this URL with the corresponding file name if appropriate.

There is a question related to SO: Secret download of a file inside the browser with the correct file name

You may also find this link useful: Filename attribute for Inline Content-Disposition Pointless?

I think you cannot download a file without streaming. You must use a stream for I / O.

Answer 2

To set the name of the download file, follow these steps for the response object in the servlet code

  response.setHeader("Content-disposition", "attachment; filename=" + "new_audio1.amr" ); 

EDIT: I see you are already doing this. Just try to remove the slashes that you added.