Develop Reference

Contents and format of URI when uploading or downloading file

I have an application which is an API on a server (say 192.168.0.2), to which files (of any format) can be uploaded or from which they can be downloaded.
If an application on another machine on the network (say 192.196.0.3) wants to upload a file to the API, it passes the information in a JSON Object e.g. { "FILE_LOCATION":"file:/192.168.0.3/c:/testDocs/testFile.docx" }
The code in the api goes roughly:
private static void doPost (HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
String errorMessage = "";
try
{
String src = request.getParameter ("src");
Object obj = jsonParser.parse (src);
JSONObject jsonObj = (JSONObject) obj;
String fileLocation = (String) jsonObj.get ("FILE_LOCATION");
URI uri = new URI (fileLocation);
URL url = uri.toURL (); // get URL from your uri object
URLConnection urlConnection = url.openConnection ();
InputStream is = urlConnection.getInputStream ();
System.out.println ("InputStream = " + is);
if (is != null)
{
// create output file, output stream etc
}
}
catch (Exception e)
{
errorMessage = e.getMessage ();
System.out.println (e.getClass().getName () + " : " + errorMessage);
}
PrintWriter pw = response.getWriter ();
pw.append (errorMessage);
}
The system log invariably shows something like:
"java.io.FileNotFoundException : 192.168.0.3/c:/testDocs/testFile.docx (No such file or directory)"
What am I doing wrong? I am convinced that the fault lies in the way I have constructed the String which will be used to create the URI.

That does not look like a valid file: URL. On Windows you could try to check if a file is accessible from a remote server if dir works in CMD.EXE. For example, try UNC pathname:
dir \\IP_OR_HOSTNAME\NAMEOFSHARE\path\etc\filename.xyz
If that works - and that rather depends on whether the remote server is serving that filesystem as NAMEOFSHARE - then the equivalent file: encoding of above would be:
String u = new File("\\\\IP_OR_HOSTNAME\\NAMEOFSHARE\\path\\etc\\filename.xyz").toURL().toString();
==> "file://IP_OR_HOSTNAME/NAMEOFSHARE/path/etc/filename.xyz"

java unknown protocol: e when downloading a file

I'm a beginner to java file handling. I tired to get a bin file (en-parser-chunking.bin) from my hard disk partition to my web application. So far I have tried below code and it gives me the output in my console below.
unknown protocol: e
these are the code samples I have tried so far
//download file
public void download(String url, File destination) throws IOException {
URL website = new URL(url);
ReadableByteChannel rbc = Channels.newChannel(website.openStream());
FileOutputStream fos = new FileOutputStream(destination);
fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);
}
public void parserAction() throws Exception {
//InputStream is = new FileInputStream("en-parser-chunking.bin");
File modelFile = new File("en-parser-chunking.bin");
if (!modelFile.exists()) {
System.out.println("Downloading model.");
download("E:\\Final Project\\Softwares and tools\\en-parser-chunking.bin", modelFile);
}
ParserModel model = new ParserModel(modelFile);
Parser parser = ParserFactory.create(model);
Parse topParses[] = ParserTool.parseLine(line, parser, 1);
for (Parse p : topParses){
//p.show();
getNounPhrases(p);
}
}
getting a file in this way is possible or I have done it wrong ?
note - I need to get this from my hard disk. not download from the internet

the correct URL for a local file is:
file://E:/Final Project/Softwares and tools/en-parser-chunking.bin
where file is the protocol.
You can also you:
new File("E:/Final Project/Softwares and tools/en-parser-chunking.bin").toURL()
to create a URL from your file.
I also recomment to use slash as file seperator instead of backslash

How to upload multiple files in Play Framework using Java

Hi i have been trying to upload image file in Play Framework. I have been trying out with Java File Upload since morning but unable to do so. I have seen [JavaFileUpload][1] tutorial available on framework website. But i am still not successful. Here is my code which i am trying to run:
Http.MultipartFormData body = request().body().asMultipartFormData();
List<Http.MultipartFormData.FilePart> fileParts = body.getFiles();
for (Http.MultipartFormData.FilePart filePart : fileParts) {
String filename = filePart.getFilename();
File file = filePart.getFile(); //error comes on this line
if (filePart.getFilename().toLowerCase().endsWith(".png")) {
//saving here but how?
} else {
return badRequest("Invalid request, only PNGs are allowed.");
}
}
but problem is that whenever i try to get the file i am having this conversion error:
java.lang.Object cannot be converted to java.io.File
Anyone can guide me in the direction? if we see the official document there is no proper documentation on how to upload multiple files. If anyone can show me some website which can helps me in that direction that will be also helpful

I'm using Play 2.4 and
FilePart filePart = request().body().asMultipartFormData()
.getFile("myFileKey");
File file = filePart.getFile();
With Play 2.2 I used for multiple file uploads:
MultipartFormData mfd = request().body().asMultipartFormData();
List<FilePart> filePartList = mfd.getFiles();
FilePart filePart = filePartList.get(0);

So after lots of trouble i was able to figure out the answer to my question. Here i am going to post the answer so it helps other people searching the answer to the same problem i faced
The controller function call which will upload the files looks like this:
Http.MultipartFormData body = request().body().asMultipartFormData();
List<Http.MultipartFormData.FilePart> fileParts = body.getFiles();
for (Http.MultipartFormData.FilePart filePart : fileParts) {
if (filePart.getFilename().toLowerCase().endsWith(".png")) {
String filename = filePart.getFilename();
Files.write(Paths.get(filename + ".png"), readContentIntoByteArray((File) filePart.getFile()));
} else {
return badRequest("Invalid request, only PNGs are allowed.");
}
}
I am using a function call to read the content of the file into byte array and save them inside the file:
private static byte[] readContentIntoByteArray(File file) {
FileInputStream fileInputStream = null;
byte[] bFile = new byte[(int) file.length()];
try {
//convert file into array of bytes
fileInputStream = new FileInputStream(file);
fileInputStream.read(bFile);
fileInputStream.close();
} catch (Exception e) {
e.printStackTrace();
}
return bFile;
}
Remember you can choose whatever the path you want to save the file at Paths.get(filename + ".png")

How to store uploaded mulitple pdf file to a specific location in java?

i want to store uploaded file in a specific location in java. if i upload a.pdf then i want it to store this at "/home/rahul/doc/upload/". i went through some questions and answers of stack overflow but i am not satisfied with solutions.
i am working with Play Framework 2.1.2. i am not working with servlet.
i am uploading but it is storing file into temp directory but i want that file store into a folder as not a temp file i want that file like a.pdf in folder not like temp file.
public static Result upload() {
MultipartFormData body = request().body().asMultipartFormData();
FilePart filePart1 = body.getFile("filePart1");
File newFile1 = new File("path in computer");
File file1 = filePart1.getFile();
InputStream isFile1 = new FileInputStream(file1);
byte[] byteFile1 = IOUtils.toByteArray(isFile1);
FileUtils.writeByteArrayToFile(newFile1, byteFile1);
isFile1.close();
}
but i am not satisfied with this solution and i am uploading multiple doc files.
for eg. i upload one doc ab.docx then after upload it is storing temp directory and file is this:
and it's location is this: /tmp/multipartBody5886394566842144137asTemporaryFile
but i want this: /upload/ab.docx
tell me some solution to fix this.

Everything's correct as a last step you need to renameTo the temporary file into your upload folder, you don't need to play around the streams it's as simple as:
public static Result upload() {
Http.MultipartFormData body = request().body().asMultipartFormData();
FilePart upload = body.getFile("picture");
if (upload != null) {
String targetPath = "/your/target/upload-dir/" + upload.getFilename();
upload.getFile().renameTo(new File(targetPath));
return ok("File saved in " + targetPath);
} else {
return badRequest("Something Wrong");
}
}
BTW you should implement some checking if targetPath doesn't exist to prevent errors and/or overwrites. Typical approach is incrementing the file name if file with the same name already exists, for an example sending a.pdf three times should save the files as a.pdf, a_01.pdf, a_02.pdf, etc.

i just completed it. My solution is working fine.
My solution of uploading multiple files is :
public static Result up() throws IOException{
MultipartFormData body = request().body().asMultipartFormData();
List<FilePart> resourceFiles=body.getFiles();
InputStream input;
OutputStream output;
File part1;
String prefix,suffix;
for (FilePart picture:resourceFiles) {
part1 =picture.getFile();
input= new FileInputStream(part1);
prefix = FilenameUtils.getBaseName(picture.getFilename());
suffix = FilenameUtils.getExtension(picture.getFilename());
part1=new File("/home/rahul/Documents/upload",prefix+"."+suffix);
part1.createNewFile();
output = new FileOutputStream(part1);
IOUtils.copy(input, output);
Logger.info("Uploaded file successfully saved in " + part1.getAbsolutePath());
}

java servlet cos multipart: save inpustream for later storage

I am using COS multipart to handle file upload on the servlet.
When processing the parts i need to rename the file with an extra posted field (ParamPart), in this case 'artikelcode' needs to be prepended to the filename.
So instead of directly writing the FilePart to disk i need to save the inputstream in memory.
This is the code i have so far:
MultipartParser multipartParser = new MultipartParser(request, 100000000);
String artikelcode = null;
String filename = null;
InputStream in = null;
while ((part = multipartParser.readNextPart()) != null) {
if (part.isFile()) {
FilePart filePart = (FilePart) part;
filename = filePart.getFileName();
//long fileSize = filePart.writeTo(new File(fileSavePath));
if (filename != null) in = filePart.getInputStream();
}
if (part.isParam()) {
ParamPart paramPart = (ParamPart) part;
if (paramPart.getName().equals("artikelcode")) artikelcode = paramPart.getStringValue();
}
}
if (in != null)
{
String fileSavePath = "c:\\upload\\"+artikelcode+"_"+filename;
File file = new File(fileSavePath);
OutputStream out = new FileOutputStream(file);
IOUtils.copy(in, out);
out.close();
}
When the file is saved on disk, it is empty!
Thanks for your help!!

Calling readNextPart() invalidates any data that you got from the previous part.
Here is a better approach: Always save the file with a temporary name and then rename it.
This allows you to handle a lot of common errors graciously like: Disk full, errors while saving, etc. because you never overwrite existing files until you are 100% sure the new file is complete.

Try to replace double slash with single slash like this..
String fileSavePath = "c:/upload/"+artikelcode+"_"+filename;