I've got a model like this:
class MyModel(models.Model):
name = models.CharField(max_length=255)
code = models.FileField()
When a new MyModel is submitted, I want to allow for the code field to be left empty, in which case I need Django to create an empty file (with arbitrary name).
Question is: what is the right way to do it?
I couldn't find anything related in the docs, so I was looking at manually editing request.FILES before feeding it t开发者_运维百科o MyModelForm(), but this looks like a dirty hack to me... Any ideas?
Thanks.
I would store the code input in a CharField and then create a separate function that accesses the model and if the code does not contain any harmful methods it is then written to a file.
This takes care of creating the file (as a blank CharField will simply be outputted to an empty file) and allows for delegation to a security checker. Your setup would then look something like the following: Model:
class MyModel(models.Model):
name = models.CharField(max_length=255)
code = models.CharField(MAX_FILE_LENGTH)
View:
def Submit_Code(request):
#Create MyModel using POST data
process_input_file(NEWLY_CREATED_MODEL_NAME)
return HttpResponse("Upload Successful")
def process_input_file(modelName):
#assuming unique name. Use "id=" instead if needed.
mm = MyModel.objects.get(name=modelName)
if passes_security_checks(mm.code):
f = open(mm.name, "r")
f.write(mm.code)
f.close()
Edit New view:
def Submit_Code(request):
mm = MyModel()
mm.name = request.POST.get('name')
f = open(mm.name,"r")
f.write(request.POST.get('code')
f.close()
#then associate the newly created file with the FileField however you want
#passing through authentication/checking if need be.
return HttpResponse("Upload Successful")
Just set the field to allow null and blank. This will make the field optional.
code = models.FileField(null=True, blank=True)
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