How to populate a field in a django model - django

Question

I have a Django project in which I have a view subclassed from the Django CreateView class. This view is used to upload a file to the server, and uses an UploadedFile model which I have created. The UploadedFile also needs to be associated with a project.
The project id is passed in as part of the URL: (r'^projects/(?P<proj_key>\d+)/$', UploadedFileCreateView.as_view(), {}, 'upload-new')
The problem is that I am not sure where the appropriate place is to associate this key with my model. Is there a method of CreateView or one of its ancestors that I should override that creates the model, or can this be done anywhere in my code in one of the methods I already override (this feels hacky though).
Furthermore, the project attribute of my UploadedFile is defined as a ForeignKey of type Project. How do I get the Project to associate with it?
Here is my model definition:
class Project(models.Model):
"""This is a project that is owned by a user and contains many UploadedFiles."""
name = models.CharField(max_length=200)
class UploadedFile(models.Model):
"""This represents a file that has been uploaded to the server."""
STATE_UPLOADED = 0
STATE_ANNOTATED = 1
STATE_PROCESSING = 2
STATE_PROCESSED = 4
STATES = (
(STATE_UPLOADED, "Uploaded"),
(STATE_ANNOTATED, "Annotated"),
(STATE_PROCESSING, "Processing"),
(STATE_PROCESSED, "Processed"),
)
status = models.SmallIntegerField(choices=STATES,
default=0, blank=True, null=True)
file = models.FileField(upload_to=settings.XML_ROOT)
project = models.ForeignKey(Project)
def __unicode__(self):
return self.file.name
def name(self):
return os.path.basename(self.file.name)
def save(self, *args, **kwargs):
if not self.status:
self.status = self.STATE_UPLOADED
super(UploadedFile, self).save(*args, **kwargs)
def delete(self, *args, **kwargs):
os.remove(self.file.path)
self.file.delete(False)
super(UploadedFile, self).delete(*args, **kwargs)
Here is my view definition:
class UploadedFileCreateView(CreateView):
model = UploadedFile
def form_valid(self, form):
logger.critical("Inside form_valid")
self.object = form.save()
f = self.request.FILES.get('file')
data = [{'name': f.name,
'url': settings.MEDIA_URL + "files/" + f.name.replace(" ", "_"),
'project': self.object.project.get().pk,
'delete_url': reverse('fileupload:upload-delete',
args=[self.object.id]),
'delete_type': "DELETE"}]
response = JSONResponse(data, {}, response_mimetype(self.request))
response['Content-Disposition'] = 'inline; filename=files.json'
return super(UploadedFileCreateView, self).form_valid(form)
def get_context_data(self, **kwargs):
context = super(UploadedFileCreateView, self).get_context_data(**kwargs)
return context

Answer 1

You could do it right where you are calling form.save(). Just pass commit=False so that it won't save it to the db until you add the project id. For example:
self.object = form.save(commit=False)
self.object.project_id = self.kwargs['proj_key']
self.object.save()
Just make sure your form excludes the project field.
EDIT: to exclude the field, add an excludes variable to the form meta class:
class UploadedFileForm(forms.ModelForm):
class Meta:
model = UploadedFile
excludes = ('project',)