问题
I'm trying to make dynamic upload path to FileField model. So when user uploads a file, Django stores it to my computer /media/(username)/(path_to_a_file)/(filename).
E.g. /media/Michael/Homeworks/Math/Week_1/questions.pdf or /media/Ernie/Fishing/Atlantic_ocean/Good_fishing_spots.txt
VIEWS
@login_required
def add_file(request, **kwargs):
if request.method == 'POST':
form = AddFile(request.POST, request.FILES)
if form.is_valid():
post = form.save(commit=False)
post.author = request.user
post.parent = Directory.objects.get(directory_path=str(kwargs['directory_path']))
post.file_path = str(kwargs['directory_path'])
post.file_content = request.FILES['file_content'] <-- need to pass dynamic file_path here
post.save()
return redirect('/home/' + str(post.author))
MODELS
class File(models.Model):
parent = models.ForeignKey(Directory, on_delete=models.CASCADE)
author = models.ForeignKey(User, on_delete=models.CASCADE)
file_name = models.CharField(max_length=100)
file_path = models.CharField(max_length=900)
file_content = models.FileField(upload_to='e.g. /username/PATH/PATH/..../')
FORMS
class AddFile(forms.ModelForm):
class Meta:
model = File
fields = ['file_name', 'file_content']
What I have found was this, but after trial and error I have not found the way to do it. So the "upload/..." would be post.file_path, which is dynamic.
def get_upload_to(instance, filename):
return 'upload/%d/%s' % (instance.profile, filename)
class Upload(models.Model):
file = models.FileField(upload_to=get_upload_to)
profile = models.ForeignKey(Profile, blank=True, null=True)
回答1:
You can use some thing like this(i used it in my project):
import os
def get_upload_path(instance, filename):
return os.path.join(
"user_%d" % instance.owner.id, "car_%s" % instance.slug, filename)
Now:
photo = models.ImageField(upload_to=get_upload_path)
回答2:
Since the file_path
is an attribute on the File
model, can you not build the full path something like this:
import os
def create_path(instance, filename):
return os.path.join(
instance.author.username,
instance.file_path,
filename
)
And then reference it from your File
model:
class File(models.Model):
...
file_content = models.FileField(upload_to=create_path)
来源:https://stackoverflow.com/questions/50591304/django-dynamic-filefield-upload-to