Newbie alert. I have a model which contains a FileField. The standard way I'd like this field to be 'filled' is in code, rather than by user interaction but there are circumstances where it needs to have a standard 'upload- from-form' behaviour.
The code method generates a file in an accessible place on the file system. My question, after some time searching Google and this group and failing to find an answer, is: how do I manipulate the FileField so that it is properly set to my existing file? That is, I want it to behave as if a user uploaded the file, except I've done it in code instead. For what it's worth, I'm using the SVN version, current as of about a week ago. Thanks in advance, Anthony. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com 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 -~----------~----~----~----~------~----~------~--~---
