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


Is it possible to change the file name of an uploaded file in django? I searched, but couldn’t find any answer.

My requirement is whenever a file is uploaded its file name should be changed in the following format.

format = userid + transaction_uuid + file_extension

Thank you very much…


How are you uploading the file?
I assume with the FileField.

The documentation for FileField.upload_to says that the upload_to field,

may also be a callable, such as a
function, which will be called to
obtain the upload path, including the
filename. This callable must be able
to accept two arguments, and return a
Unix-style path (with forward slashes)
to be passed along to the storage
system. The two arguments that will be
passed are:

"instance": An instance of
the model where the FileField is
defined. More specifically, this is
the particular instance where the
current file is being attached.

"filename":The filename that was
originally given to the file. This may
or may not be taken into account when
determining the final destination

So it looks like you just need to make a function to do your name handling and return the path.

def update_filename(instance, filename):
    path = "upload/path/"
    format = instance.userid + instance.transaction_uuid + instance.file_extension
    return os.path.join(path, format)
Answered By: monkut

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]


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" % (,)
        # get filename
            complaint_id = "cid_%s" % (,)
            filename = '{}.{}.{}.{}'.format(prefix, project, complaint_id, ext)
            # 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.

Answered By: A.J.

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"))
Answered By: nicolas

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))
Answered By: Bárbara Este
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 =

    path = "uploads/{}/{}/{}/".format(todays_date.year, todays_date.month,
    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)
Answered By: Haykins

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


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

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

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

Answered By: Aniket Pawar

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
, 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(,filename.split('.')[-1])

 avatar = models.ImageField(_("avatar"),upload_to=AvatarSave)
Answered By: Oliver Sitán