I think this is trivial considering that the example does not use request.POST['title'] They must of just forgot to use the title field in the view?
One practical example could be to use the title as the filename which you intend to store in your filesystem. You can do this by changing the filename of the file object to the "title" string you stored in the request. This would be done in a view method. -Rob On Thu, Jan 27, 2011 at 10:53 AM, hank23 <[email protected]> wrote: > I'm trying to code and implement a file upload screen. I've looked at > the documentation at: > > http://docs.djangoproject.com/en/1.2/topics/http/file-uploads/ > > and the document shows a simple upload form called UploadFileForm > which is shown with two fields coded for it. I'm assuming that the > field named "file" of type forms.FileField is just the file > itself(location, name, and data), but if so what's the "title" field > of type forms.CharField used for and how is it processed? I don't > think it's mentioned at all in the documentation except for being > shown to be part of the form. Can someone please explain its use and > how it should be processed? Thanks. > > -- > You received this message because you are subscribed to the Google Groups > "Django users" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]<django-users%[email protected]> > . > For more options, visit this group at > http://groups.google.com/group/django-users?hl=en. > > -- Robert McQueen Massachusetts Institute of Technology Class of 2012, Course 6.3 - Computer Science [email protected] -- You received this message because you are subscribed to the Google Groups "Django users" 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/django-users?hl=en.
