Issue
In my Django app, I would like to define upload path and file dynamically when saving to AWS S3. I am able to so far save to S3 directly the file however, I want to set the path and filename it self.
So for example, upon upload i want it to be in the S3 path bucketname\employeeid\file_randomnumber.png
How can i do something like this ?
Below are my code:
https://gitlab.com/firdausmah/railercom/blob/master/railercomapp/api.py
@api_view(['POST'])
def update_employee_image(request):
# ----- YAML below for Swagger -----
"""
description: update employee image.
parameters:
- name: employee_id
type: integer
required: true
location: form
- name: face_image
type: file
required: true
location: form
"""
parser_classes = (FileUploadParser,)
employee_id = request.POST['employee_id']
face_image_obj = request.data['face_image']
employee = Employee.objects.get(id = employee_id)
logging.debug(f"API employee username {employee.username}")
#employee.face_image = face_image_obj
employee.upload = face_image_obj <--- here is where it assign the file to S3
employee.save()
return Response("Employee Updated!", status=status.HTTP_200_OK)
https://gitlab.com/firdausmah/railercom/blob/master/railercomapp/models.py
class Employee(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, related_name='employee')
company = models.ForeignKey(Company)
username = models.CharField(max_length=30, blank=False)
upload = models.FileField(blank=True) <--- S3 field
https://gitlab.com/firdausmah/railercom/blob/master/railercom/settings.py (AWS settings)
AWS_ACCESS_KEY_ID = config('AWS_ACCESS_KEY_ID')
AWS_SECRET_ACCESS_KEY = config('AWS_SECRET_ACCESS_KEY')
AWS_STORAGE_BUCKET_NAME = config('AWS_STORAGE_BUCKET_NAME')
AWS_S3_CUSTOM_DOMAIN = '%s.s3.amazonaws.com' % AWS_STORAGE_BUCKET_NAME
AWS_S3_OBJECT_PARAMETERS = {
'CacheControl': 'max-age=86400',
}
DEFAULT_FILE_STORAGE = 'railercomapp.storage_backends.MediaStorage'
https://gitlab.com/firdausmah/railercom/blob/master/railercomapp/storage_backends.py from storages.backends.s3boto3 import S3Boto3Storage
class MediaStorage(S3Boto3Storage):
location = 'media/yy'
file_overwrite = False
I have based my django aws S3 solution on https://simpleisbetterthancomplex.com/tutorial/2017/08/01/how-to-setup-amazon-s3-in-a-django-project.html.
Solution
You can override the storage get_available_name
method.
Here's an example. Modify to get the exact filename scheme you want.
class MediaStorage(S3Boto3Storage):
location = 'media/yy'
file_overwrite = False
def get_available_name(self, name, max_length=None):
custom_name = f'/employeeid/{name}_randomnumber.png'
return super().get_available_name(custom_name, max_length)
Docs: Writing a custom storage system
Answered By - Håken Lid
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