[Zope] File upload problem?

[Roger Ineichen]

Hello together, 
I've a problem with upload a picture and the method cookId(id='',
title='', file=file). This method doesn't work for me? The method
cookId() doesn't give a id or a title back. Why?
Whats wrong?

Form
--------------------------
<form action="addFile" method="post" enctype="multipart/form-data">
<input type="file" name="file">
<input type="submit" value="upload">
</form>


Python  (not pythonScript)
--------------------------
def addFile(self, file):
  "This method adds a file"
  # create the file
  if(file):
    id, title = cookId(id='', title='', file=file)
    self.manage_addProduct['P01_DownloadFolder'].manage_addFile(id=id,
title='', file=file)

Second Way I tryed (the manage_addFile() call the cookId by standard.
--------------------------
def addFile(self, file):
  "This method adds a file"
  # create the file
  self.manage_addProduct['P01_DownloadFolder'].manage_addFile(id='',
title='', file=file)


cookId is a method from Image.py
--------------------------------
def cookId(id, title, file):
  """ it gets the id from a file """
  if not id and hasattr(file,'filename'):
    filename=file.filename
    title=title or filename
    id=filename[max(string.rfind(filename, '/'),
                    string.rfind(filename, '\\'),
                    string.rfind(filename, ':'),
               	  )+1:]
return id, title

The error: 
----------
The "id" is not defined cookId does not set the id, 
How can id get the id from the file in a different way.

Is the Form 
enctype="multipart/form-data" wrong? What should i use?


Thanks to all and have a nice week.
Roger Ineichen
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