Why are you looking up the client filename? Do you have access the the
remote computer your user is on?
Martijn
On Thu, May 14, 2009 at 10:53 AM, Stefan Droog <sdr...@educator.eu> wrote:
> Hi all,
>
> Currently I want to upload a file via FileUploadField. However when I select
> a file and press upload I get a FileNotFoundException (System cannot find the
> file specified). File.getCanonicalPath and file.getAbsolutePath both returns
> the wrong location.
>
> Somebody a clue why?
>
> Regards,
> Stefan
>
> Java:
>
> final FileUploadField fileUploadField = new FileUploadField("fileInput2");
>
> Form form = new Form("ajax-simpleUpload2"){
>
> private static final long
> serialVersionUID = -2623469756422643826L;
>
> �...@override
> protected void onSubmit() {
> final
> FileUpload upload = fileUploadField.getFileUpload();
>
> File file =
> new File(upload.getClientFileName());
>
> try {
>
> file.getCanonicalPath(); //Returns wrong path
>
> file.getAbsolutePath(); //Returns wrong path
>
> InputStream in = new FileInputStream(file);
> } catch
> (FileNotFoundException e) {
>
> e.printStackTrace();
> } catch
> (IOException e) {
>
> e.printStackTrace();
> }
>
>
> super.onSubmit();
> }};
> form.setMultiPart(true);
>
> form.add(fileUploadField);
> add(form);
>
>
> HTML:
>
> <form wicket:id="ajax-simpleUpload2">
> <fieldset>
> <legend>Upload
> form</legend>
> <p>
> <label
> for="upload">File</label>
> <input wicket:id="fileInput2"
> id="upload" type="file"/>
> </p>
> <input type="submit"
> value="Upload!"/>
> </fieldset>
> </form>
>
>
>
>
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