Develop Reference

In Groovy, how to properly get the file from HttpServletRequest

I am writing a REST API in Groovy script that will receive a file upload from client side.
The REST API will receive the file via HttpServletRequest.
I am trying to get the file from HttpServletRequest by getting its InputStream, then convert it to File to save to proper folder.
My code is as below:
RestApiResponse doHandle(HttpServletRequest request, RestApiResponseBuilder apiResponseBuilder, RestAPIContext context) {
InputStream inputStream = request.getInputStream()
def file = new File(tempFolder + "//" + fileName)
FileOutputStream outputStream = null
try
{
outputStream = new FileOutputStream(file, false)
int read;
byte[] bytes = new byte[DEFAULT_BUFFER_SIZE];
while ((read = inputStream.read(bytes)) != -1) {
outputStream.write(bytes, 0, read);
}
}
finally {
if (outputStream != null) {
outputStream.close();
}
}
inputStream.close();
// the rest of the code
}
The files are created, but all of them are corrupted.
When I try to open them with Notepad, all of them have, at the beginning, some thing similar to the below:
-----------------------------134303111730200325402357640857
Content-Disposition: form-data; name="pbUpload1"; filename="Book1.xlsx"
Content-Type: application/vnd.openxmlformats-officedocument.spreadsheetml.sheet
Am I doing this wrong? How do I get the file correctly?

Found the solution with MultipartStream
import org.apache.commons.fileupload.MultipartStream
import org.apache.commons.io.FileUtils
InputStream inputStream = request.getInputStream()
//file << inputStream;
String fileName = "";
final String CD = "Content-Disposition: "
MultipartStream multipartStream = new MultipartStream(inputStream, boundary);
//Block below line because it always return false for some reason
// but should be used as stated in document
//boolean nextPart = multipartStream.skipPreamble();
//Block below line as in my case, the part I need is at the first part
// or maybe I should use it and break after successfully get the file name
//while(nextPart) {
String[] headers = multipartStream.readHeaders().split("\\r\\n")
ContentDisposition cd = null
for (String h in headers) {
if (h.startsWith(CD)) {
cd = new ContentDisposition(h.substring(CD.length()));
fileName = cd.getParameter("filename"); }
}
def file = new File(tempFolder + "//" + fileName)
ByteArrayOutputStream output = new ByteArrayOutputStream(1024)
try
{
multipartStream.readBodyData(output)
FileUtils.writeByteArrayToFile(file, output.toByteArray());
}
finally {
if (output != null) {
output.flush();
output.close();
}
}
inputStream.close();

Android java unable to download file

am trying to download file from server , but no success my code seems to be ok
URL url = null;
URLConnection con = null;
int i;
try {
url = new URL(downlink); // url : http://10.0.2.2:800/myproject/down/file9.txt
con = url.openConnection();
String dest_path = c.getFilesDir().getPath() + "/textfile.txt"; //Download Location set to : /data/data/com.myproject.androidt/files/textfile.txt
File file = new File(dest_path);
BufferedInputStream bis = new BufferedInputStream(con.getInputStream());
FileOutputStream fos = context.openFileOutput("textfile.txt", Context.MODE_PRIVATE);
BufferedOutputStream bos = new BufferedOutputStream(fos);
while ((i = bis.read()) != -1) {
bos.write(i);
}
bos.flush();
bis.close();
return true;
} catch (MalformedInputException malformedInputException) {
Log.d("dark","Failure : MalformedInputException occured in downloading");
// error in download
return false;
} catch (IOException ioException) {
Log.d("dark","Failure : IO Error occured in downloading");
return false;
// error in download
}
so please help i get IO exception , do not know what's wrong with code :(

I believe you have to call the openFileOutput method to get the FileOutputStream
FileOutputStream fos = openFileOutput("textfile.txt", Context.MODE_PRIVATE);
and you only need the file name not the path.

How to reference a File in raw folder in Android

I just want to create a File object like this
File myImageFile = new File ("image1") ;
but it is giving me exception of FileNotFoundException
How can i reference a file inside my raw Folder
EDIT:
Actually i wanted to do something like this
MultipartEntity multipartEntity= new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
multipartEntity.addPart("uploaded", new FileBody(new File("myimage")));

Generally you access the files through getResources().openRawResource(R.id._your_id). If you absolutely need a File reference to it, one option is to copy it over to internal storage:
File file = new File(this.getFilesDir() + File.separator + "DefaultProperties.xml");
try {
InputStream inputStream = resources.openRawResource(R.id._your_id);
FileOutputStream fileOutputStream = new FileOutputStream(file);
byte buf[]=new byte[1024];
int len;
while((len=inputStream.read(buf))>0) {
fileOutputStream.write(buf,0,len);
}
fileOutputStream.close();
inputStream.close();
} catch (IOException e1) {}
Now you have a File that you can access anywhere you need it.

here are 2 functions. one to read from RAW and one from the Assets
/**
* Method to read in a text file placed in the res/raw directory of the
* application. The method reads in all lines of the file sequentially.
*/
public static void readRaw(Context ctx,int res_id) {
InputStream is = ctx.getResources().openRawResource(res_id);
InputStreamReader isr = new InputStreamReader(is);
BufferedReader br = new BufferedReader(isr, 8192); // 2nd arg is buffer
// size
// More efficient (less readable) implementation of above is the
// composite expression
/*
* BufferedReader br = new BufferedReader(new InputStreamReader(
* this.getResources().openRawResource(R.raw.textfile)), 8192);
*/
try {
String test;
while (true) {
test = br.readLine();
// readLine() returns null if no more lines in the file
if (test == null)
break;
}
isr.close();
is.close();
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
and from Assets folder
/**
* Read a file from assets
*
* #return the string from assets
*/
public static String getQuestions(Context ctx,String file_name) {
AssetManager assetManager = ctx.getAssets();
ByteArrayOutputStream outputStream = null;
InputStream inputStream = null;
try {
inputStream = assetManager.open(file_name);
outputStream = new ByteArrayOutputStream();
byte buf[] = new byte[1024];
int len;
try {
while ((len = inputStream.read(buf)) != -1) {
outputStream.write(buf, 0, len);
}
outputStream.close();
inputStream.close();
} catch (IOException e) {
}
} catch (IOException e) {
}
return outputStream.toString();
}

You can open it as InputStream, I don't know if possible as a file:
int rid = resources.getIdentifier(packageName + ":raw/" + fileName, null, null);
//get the file as a stream
InputStrea is = resources.openRawResource(rid);

You can use InputStreamBody instead of FileBody so you can use it like this:
InputStream inputStream = resources.openRawResource(R.raw.yourresource);
MultipartEntity multipartEntity= new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
multipartEntity.addPart("uploaded", new InputStreamBody(inputStream));

Download a file through an HTTP Get in java

I've written a download Servlet to return a file based on a messageID parameter. Below 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();
}
I then wrote some code to retrieve 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 works fine. I can download the audio file and it works correctly. What I want to know is how to, if possible, get the name of the file as it is downloaded instead of giving it my own name. Also, is it possible to get the file without having to read from the stream (maybe some library that does it for you)? I kind of want to hide the dirty stuff.
Thanks

For setting the download file name do the following on response object in Servlet code
response.setHeader("Content-disposition",
"attachment; filename=" +
"new_audio1.amr" );
EDIT:
I see you are already doing it. Just try removing the slashes you have added.

With attachment, the file will be served with the provided name properly. When inline, browsers seem to ignore filename, and usually give the servletname part of the URL as default name when saving the inline contents.
You could try mapping that URL to an appropriate filename, if that is suitable.
Here's a SO related question: Securly download file inside browser with correct filename
You may also find this link useful: Filename attribute for Inline Content-Disposition Meaningless?
I think you cannot download file without streaming. For I/O you must use stream.

file upload using java servlet as a service without a web browser

I am very new to java and servlet programming.
I am not sure whether it is possible to write a servlet which when passed a URL from the local client machine, uploads the file to the server.
basically on the client machine we have a C# program and on the server side we have Apache-tomcat installed. I need to upload file(s) to the server using C# program on client machine.
Should I provide any more information (?)
Thanks in Advance

Note this code illustrates the general idea and not guaranteed to work without modification.
The C# file upload part
// this code shows you how the browsers wrap the file upload request, you still can fine a way simpler code to do the same thing.
public void PostMultipleFiles(string url, string[] files)
{
string boundary = "----------------------------" + DateTime.Now.Ticks.ToString("x");
HttpWebRequest httpWebRequest = (HttpWebRequest)WebRequest.Create(url);
httpWebRequest.ContentType = "multipart/form-data; boundary=" + boundary;
httpWebRequest.Method = "POST";
httpWebRequest.KeepAlive = true;
httpWebRequest.Credentials = System.Net.CredentialCache.DefaultCredentials;
Stream memStream = new System.IO.MemoryStream();
byte[] boundarybytes =System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary +"\r\n");
string formdataTemplate = "\r\n--" + boundary + "\r\nContent-Disposition: form-data; name=\"{0}\";\r\n\r\n{1}";
string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\n Content-Type: application/octet-stream\r\n\r\n";
memStream.Write(boundarybytes, 0, boundarybytes.Length);
for (int i = 0; i < files.Length; i++)
{
string header = string.Format(headerTemplate, "file" + i, files[i]);
//string header = string.Format(headerTemplate, "uplTheFile", files[i]);
byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
memStream.Write(headerbytes, 0, headerbytes.Length);
FileStream fileStream = new FileStream(files[i], FileMode.Open,
FileAccess.Read);
byte[] buffer = new byte[1024];
int bytesRead = 0;
while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
{
memStream.Write(buffer, 0, bytesRead);
}
memStream.Write(boundarybytes, 0, boundarybytes.Length);
fileStream.Close();
}
httpWebRequest.ContentLength = memStream.Length;
Stream requestStream = httpWebRequest.GetRequestStream();
memStream.Position = 0;
byte[] tempBuffer = new byte[memStream.Length];
memStream.Read(tempBuffer, 0, tempBuffer.Length);
memStream.Close();
requestStream.Write(tempBuffer, 0, tempBuffer.Length);
requestStream.Close();
try
{
WebResponse webResponse = httpWebRequest.GetResponse();
Stream stream = webResponse.GetResponseStream();
StreamReader reader = new StreamReader(stream);
string var = reader.ReadToEnd();
}
catch (Exception ex)
{
response.InnerHtml = ex.Message;
}
httpWebRequest = null;
}
and to understand how the above code was written you might wanna take a look at How does HTTP file upload work?
POST /upload?upload_progress_id=12344 HTTP/1.1
Host: localhost:3000
Content-Length: 1325
Origin: http://localhost:3000
... other headers ...
Content-Type: multipart/form-data; boundary=----WebKitFormBoundaryePkpFF7tjBAqx29L
------WebKitFormBoundaryePkpFF7tjBAqx29L
Content-Disposition: form-data; name="MAX_FILE_SIZE"
100000
------WebKitFormBoundaryePkpFF7tjBAqx29L
Content-Disposition: form-data; name="uploadedfile"; filename="hello.o"
Content-Type: application/x-object
... contents of file goes here ...
------WebKitFormBoundaryePkpFF7tjBAqx29L--
and finally all you have to do is to implement a servlet that can handle the file upload request, then you do whatever that you want to do with the file, take a look at this file upload tutorial
protected void processRequest(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
// Create path components to save the file
final String path = request.getParameter("destination");
final Part filePart = request.getPart("file");
final String fileName = getFileName(filePart);
OutputStream out = null;
InputStream filecontent = null;
final PrintWriter writer = response.getWriter();
try {
out = new FileOutputStream(new File(path + File.separator
+ fileName));
filecontent = filePart.getInputStream();
int read = 0;
final byte[] bytes = new byte[1024];
while ((read = filecontent.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
writer.println("New file " + fileName + " created at " + path);
LOGGER.log(Level.INFO, "File{0}being uploaded to {1}",
new Object[]{fileName, path});
} catch (FileNotFoundException fne) {
writer.println("You either did not specify a file to upload or are "
+ "trying to upload a file to a protected or nonexistent "
+ "location.");
writer.println("<br/> ERROR: " + fne.getMessage());
LOGGER.log(Level.SEVERE, "Problems during file upload. Error: {0}",
new Object[]{fne.getMessage()});
} finally {
if (out != null) {
out.close();
}
if (filecontent != null) {
filecontent.close();
}
if (writer != null) {
writer.close();
}
}
}
private String getFileName(final Part part) {
final String partHeader = part.getHeader("content-disposition");
LOGGER.log(Level.INFO, "Part Header = {0}", partHeader);
for (String content : part.getHeader("content-disposition").split(";")) {
if (content.trim().startsWith("filename")) {
return content.substring(
content.indexOf('=') + 1).trim().replace("\"", "");
}
}
return null;
}