[Zope] File uploading

[Pawel Lewicki]

> Pawel Lewicki wrote:
> > Hallo,
> > I have 2 forms in my CMF site (but I suppose that is not a problem).
Both of
> > them have
> >  <input name="image_file" type="file" >
> > field. I try to get the file name to add a file(image) with proper id.
> > When I access str(REQUEST.image_file) in my python script one time I get
the
> > full path to the file and the second time:
> > <ZPublisher.HTTPRequest.FileUpload instance at 020CF314>
> > The first one is better for me for explained reason but the files seem
not
> > to be uploading. So it is probably just a name not a file.
> > The first question is where is the difference? It's probably in the code
but
> > I can't find it. I don't know much about HTTP protocol but the first
thing
> > to check was the type of forms - both are "POST" type.
> > And the second - how to extract the name of the file and have it
uploaded in
> > the same time?
> >
> > Pawel Lewicki
>
>
> your_target_to_add_image.manage_addImage('',image_file)
> should do the trick, it causes zope to construct the id itself (see
> OFS/Image.py, the method is called CookId) from the filename.
>
> Additionally, image_file.filename gives you the filename of the uploaded
> file, but you might have to strip off the leading path.
>
>
> HTH,
> oliver

I have to use the second solution as context.invokeFactory(id='image_name',
type_name='Image', file=request.image_file) requires id to be given. But
image_file.filename works great.
Thanx

Pawel