dynamic upload field arguments

Question

if I have basic model for keeping track of client art:

class ArtEntry(models.Model):
    client = models.CharField(max_length=50, choices=CLIENT_CHOICES)
    job_number = models.CharField(max_length=30, unique=False, blank=False, null=False)
    filename = models.CharField(max_length=64, unique=False, blank=False, null=False)
    descriptor = models.CharField(max_length=64, unique=False, blank=True, null=True)
    date = models.DateField(("Date"), default=datetime.date.today)
    post_type = models.CharField(max_length=64, choices=POST_CHOICES)

and the last field of this model is a file upload field:

    upload = models.FileField(upload_to='documents/*/**/***')

is it possible to populate the "upload_to" argument dynamically, so that * = the input 'client' data and ** = the input 'job_number' and * = the 'post_type' choice?

possible duplicate of [USe dynamic destination folder for uploaded file in Django](http://stackoverflow.com/questions/1936614/use-dynamic-destination-folder-for-uploaded-file-in-django) — Timmy O'Mahony, May 17 '12 at 21:44

score 1 · Accepted Answer · edited May 23 '17 at 12:27

1

Sure. Just check the docs for upload_to:

This may also be a callable, such as a function, which will be called to obtain the upload path, including the filename.

Check this answer to see an example which you can easily adapt to your problem.

edited May 23 '17 at 12:27

Community

1
1

answered May 17 '12 at 21:44

arie

18,737
5
70
76

The example was EXACTLY what I needed. Configured and tested. I'm a happy camper. Thanx. – kjarsenal May 17 '12 at 22:11

score 0 · Answer 2 · answered May 17 '12 at 21:46

One Example;

def image_path(self, uploaded_file_name):
    prefix = 'documents/'
    extension = os.path.splitext(uploaded_file_name)[-1]
    if self.pk != None:
        return prefix + str(self.pk) + extension
    else:
        tmp_name = str(uuid.uuid4())
        self.temp_image = prefix + tmp_name + extension
        return self.temp_image

image_upload = models.ImageField(upload_to=image_path, null=True, blank=True)

You should use python functions.

dynamic upload field arguments

2 Answers2

Linked