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Here is my code (servlet).

boolean isMultiPart = ServletFileUpload.isMultipartContent(request);
                    if (isMultiPart) {
                        ServletFileUpload upload = new ServletFileUpload();
                        try {
                            FileItemIterator itr = upload.getItemIterator(request);
                            while (itr.hasNext()) {
                                FileItemStream item = itr.next();
                                FILE_NAME = item.getName();

                                InputStream stream = item.openStream();
                                if (item.isFormField()) {
                                    String value = Streams.asString(stream, "UTF-8");
                                }
                            }
                            try {
                                FileInputStream excelFile = new FileInputStream(new File(FILE_NAME));
                                Workbook workbook = new XSSFWorkbook(excelFile);
                                Sheet datatypeSheet = workbook.getSheetAt(0);
                                Iterator<Row> iterator = datatypeSheet.iterator();
                            } catch (Exception adfasdasd) {
                                adfasdasd.printStackTrace();
                            }

                        } catch (FileUploadException ex) {
//                            Logger.getLogger(PropertyController.class.getName()).log(Level.SEVERE, null, ex);
                        } catch (IOException exp) {
//                            Logger.getLogger(PropertyController.class.getName()).log(Level.SEVERE, null, exp);
                        }

FileInputStream excelFile = new FileInputStream(new File(FILE_NAME)); to do this i need the full path of the uploaded file. how do i get it?

Neshan Manilka
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1 Answers1

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Use getPart method of HttpServletRequest. Tutorial is at http://docs.oracle.com/javaee/6/tutorial/doc/glraq.html

Jay Smith
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