[Django]-How to change the file name of an uploaded file in Django?

You need to have a FileField with the upload_to that calls to a callback, see [1]

Your callback should call a wrapper method which gets an instance as one of the params and filename as the other. [2]

Change it the way you like and return the new path [3]

1. LOGIC

FileField(..., upload_to=method_call(params),....)

2. define method

def method_call(params):
    return u'abc'

3. Wrapper:

def wrapper(instance, filename):
    return method

this is the rapper method that you need for getting the instance.

def wrapper(instance, filename):
... Your logic
...
return wrapper

Complete Code

def path_and_rename(path, prefix):
    def wrapper(instance, filename):
        ext = filename.split('.')[-1]
        project = "pid_%s" % (instance.project.id,)
        # get filename
        if instance.pk:
            complaint_id = "cid_%s" % (instance.pk,)
            filename = '{}.{}.{}.{}'.format(prefix, project, complaint_id, ext)
        else:
            # set filename as random string
            random_id = "rid_%s" % (uuid4().hex,)
            filename = '{}.{}.{}.{}'.format(prefix, project, random_id, ext)
            # return the whole path to the file
        return os.path.join(path, filename)

    return wrapper

Call to Method

sales_attach = models.FileField("Attachment", upload_to=path_and_rename("complaint_files", 'sales'), max_length=500,
                                help_text="Browse a file")

Hope this helps.
Thanks.

if you want your function re-usable:

import hashlib
import datetime
import os
from functools import partial

def _update_filename(instance, filename, path):
    path = path

    filename = "..."

    return os.path.join(path, filename)

def upload_to(path):
    return partial(_update_filename, path=path)

You just have to use it this way:

document = models.FileField(upload_to=upload_to("my/path"))

import random
import os
def generate_unique_name(path):
    def wrapper(instance, filename):
        extension = "." + filename.split('.')[-1]
        filename = str(random.randint(10,99)) + str(random.randint(10,99)) + str(random.randint(10,99)) + str(random.randint(10,99))  + extension
        return os.path.join(path, filename)
    return wrapper

#You just have to use it this way:#

 photo = models.FileField("Attachment", upload_to=generate_unique_name("pics"),max_length=500,help_text="Browse a photo")

Incase this may help anyone.

import os
import uuid
import random

from datetime import datetime 

def user_directory_path(instance, filename):
    # Get Current Date
    todays_date = datetime.now()

    path = "uploads/{}/{}/{}/".format(todays_date.year, todays_date.month, todays_date.day)
    extension = "." + filename.split('.')[-1]
    stringId = str(uuid.uuid4())
    randInt = str(random.randint(10, 99))

    # Filename reformat
    filename_reformat = stringId + randInt + extension

    return os.path.join(path, filename_reformat)


class MyModel(models.Model):
    upload = models.FileField(upload_to=user_directory_path)

That much code is not needed you can just use single line code
fille._name=userid + transaction_uuid + file_extension

Like

class xyz(models.Model):
   file = models.FileField(upload_to="notice/")

   def add(request):
      file = request.POST['file']
      file._name = request.user.id + transaction_uuid +"."+ file._name.split('.')[1]

you can overwrite file name by overwriting _name value of file object.

You can try this also work, name_of_submission is any column you want to attach the file

import os

def update_filename(instance, filename):
    path = "upload/path/"
    ext = filename.split('.')[-1]
    format = instance.name_of_submission +    str(instance.id_application) + "." + ext
    print(format)
    return os.path.join(path, format)

The basic way is

import os

os.rename('a.txt', 'b.html')

For your situation, it would probably look like

os.rename ("a.txt", "{id}{uuid}.{ext}".format(id=userid, uuid=transaction_uuid, ext=file_extension))

Hi, i check all the answers, but someone are not updated, this is how
in 2022 works whith clean code and following the Django Documentation
Here, remember that you need to make a MIGRATION to make this work:

 def AvatarSave(instance, filename):
    #this line changes the name of the file to the user name and put the file extension at the end after the point
    return 'users/avatars/{0}.{1}'.format(instance.id,filename.split('.')[-1])

 avatar = models.ImageField(_("avatar"),upload_to=AvatarSave)