Hi twittwit, And thanks for your reply. What I want is quite simple. I want to be able to place a button in my UI. When he user clicks on this button he can retrieve a file on his machine and clicking another button it will upload it to the server hosting the website. I tried with fileupload from gwt, gwt-ext, and others but I usually get the error that my server code is not found when I click on the upload button. Would it be possible to know step by step how should I proceed?
On Aug 2, 10:09 pm, twittwit <[email protected]> wrote: > if you mean common fileupload. it should be at the server. > nope for web.xml. and nope for project.gwt.xml --> since you putting > common fileupload to the server. > try to be more clear ini what u want? and where is the problem. > > On Jul 21, 4:13 pm, Vinz369 <[email protected]> wrote: > > > > > Hello twittwit and others, > > > I've been trying to implement fileuploadfor days in my application. > > It may looks stupid but I really don't understand how it works. > > I have few questions: > > - If I want to use the same code as twittwit where should I place it? > > in my client folder or server folder? should I add something else in > > my web.xml and project.gwt.xml files? > > > I'm completely lost, please help me! > > > On Jul 18, 10:15 am, twittwit <[email protected]> wrote: > > > > perfect! thanks Manuel! > > > common-fileupload is great! > > > On Jul 18, 8:51 am, Manuel Carrasco <[email protected]> > > > wrote: > > > > > filename = item.getName(); > > > > > On Sat, Jul 18, 2009 at 12:27 AM, twittwit <[email protected]> wrote: > > > > > > ok thank you. i found the answer: > > > > > > public class MyFormHandler extends HttpServlet{ > > > > > public void doPost(HttpServletRequest request, HttpServletResponse > > > > > response) throws ServletException, IOException { > > > > > ServletFileUploadupload= new ServletFileUpload(); > > > > > > try{ > > > > > FileItemIterator iter =upload.getItemIterator(request); > > > > > > while (iter.hasNext()) { > > > > > FileItemStream item = iter.next(); > > > > > > String name = item.getFieldName(); > > > > > InputStream stream = item.openStream(); > > > > > > // Process the input stream > > > > > FileOutputStream out = new FileOutputStream > > > > > ("example.csv"); > > > > > //ByteArrayOutputStream out = new ByteArrayOutputStream > > > > > (); > > > > > int len; > > > > > byte[] buffer = new byte[8192]; > > > > > while ((len = stream.read(buffer, 0, buffer.length)) ! > > > > > = -1) { > > > > > out.write(buffer, 0, len); > > > > > } > > > > > //....... > > > > > > } > > > > > } > > > > > catch(Exception e){ > > > > > e.printStackTrace(); > > > > > } > > > > > > } > > > > > > } > > > > > > however, how can i extract the name of the csv file(client side) so > > > > > that the csv file in my server can have the same name? > > > > > thanks!! > > > > > > On Jul 18, 12:19 am, Manuel Carrasco <[email protected]> > > > > > wrote: > > > > > > In the dialog between the browser and the server, the client sends a > > > > > > multipart/form-data request and there is more information besides > > > > > > the > > > > > file > > > > > > content, like form elements values, boundary tags, etc. > > > > > > > I recommend you to use apache commons-fileupload library to handle > > > > > > multipart/form-data request in your servlets. > > > > > > > On Fri, Jul 17, 2009 at 11:44 PM, imgnik <[email protected]> wrote: > > > > > > > > hi all, > > > > > > > > i posted a question about fileupload here > > > > > > > >http://groups.google.com/group/Google-Web-Toolkit/browse_thread/threa. > > > > > .. > > > > > > > but i think i didn't phrase my question correctly. so gonna do > > > > > > > another > > > > > > > attempt. > > > > > > > > I tried to use agwtfileupload widget to send a csv file to the > > > > > > > server. and at the server i will write it as a file (for other > > > > > > > usage) > > > > > > > it by doing : (where request is the httpservletrequest and bw is > > > > > > > the > > > > > > > bufferedwriter) > > > > > > > > BufferedReader r = request.getReader(); > > > > > > > while((thisread= r.readLine())!=null){ > > > > > > > > bw.write(thisread); > > > > > > > > } > > > > > > > bw.close(); > > > > > > > } > > > > > > > catch(Exception e1){} > > > > > > > > however, i realised that the csv file send out contains : > > > > > > > > ------WebKitFormBoundaryN8Z6DOy7DqEWTwtLContent-Disposition: form- > > > > > > > data; name="uploadFormElement"; filename="first.csv"Content-Type: > > > > > > > application/octet-streamBank > > > > > > > > which results in the file created not the same as the fileupload. > > > > > > > I might be wrong in my analysis. can someone advice me? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Google Web Toolkit" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/Google-Web-Toolkit?hl=en -~----------~----~----~----~------~----~------~--~---
