how to save files to s3 buckets - django

Question

I am trying to create a barcode image file that will save to a path in my aws s3 bucket. I don't know how to link to it. My media and static files are already on aws and are working perfectly but I don't know how to set the path for this barcodemaker function to save to the aws s3 bucket.
Thank you for your patience and guidance.
my barcode function in my view
def barcodemaker():
barcodemodel = apps.get_model('barcoder', 'barcodeModel')
employee = apps.get_model('employees', 'Employee')
data = employee.objects.filter(id=1)
try:
data2 = barcodemodel.objects.latest('id')
except:
data2 = 1002390000
naa = str(data2)
naa = int(naa[-10:])
for i in data:
id_name= str(i.id)
naa += random.randint(500, 900)
mocode = 'M-'+ id_name + '-'+ str(naa)
b = barcodemodel(barcode_num=str(mocode))
b.save()
path = (>>>PATH to aws<<<,'static','media','barcodes', mocode+'.png')
with open(path, 'wb') as f:
Code128(mocode, writer=ImageWriter()).write(f)
barcode_context = {
'mocode':mocode, 'f':f
}
return barcode_context
My Static file settings in my settings file
STATIC_URL = '/static/'
STATICFILES_DIRS = [
os.path.join(BASE_DIR,"static")
]
STATIC_ROOT = os.path.join(BASE_DIR, 'staticfiles')
MEDIA_URL = '/profile_image/'
MEDIA_ROOT = os.path.join(BASE_DIR, 'static/images')
#S3 BUCKETS CONFIG
# S3 logins Data
AWS_ACCESS_KEY_ID = os.environ.get('AWS_ACCESS_KEY_ID')
AWS_SECRET_ACCESS_KEY = os.environ.get('AWS_SECRET_ACCESS_KEY')
AWS_STORAGE_BUCKET_NAME = os.environ.get('AWS_STORAGE_BUCKET_NAME')
AWS_S3_FILE_OVERWRITE = False
AWS_DEFAULT_ACL = None
DEFAULT_FILE_STORAGE = 'storages.backends.s3boto3.S3Boto3Storage'
STATICFILES_STORAGE = 'storages.backends.s3boto3.S3Boto3Storage'

Answer 1

From the documentation:
import logging
import boto3
from botocore.exceptions import ClientError
def upload_file(file_name, bucket, object_name=None):
"""Upload a file to an S3 bucket
:param file_name: File to upload
:param bucket: Bucket to upload to
:param object_name: S3 object name. If not specified then file_name is used
:return: True if file was uploaded, else False
"""
# If S3 object_name was not specified, use file_name
if object_name is None:
object_name = file_name
# Upload the file
s3_client = boto3.client('s3')
try:
response = s3_client.upload_file(file_name, bucket, object_name)
except ClientError as e:
logging.error(e)
return False
return True
then (here f is in binary mode):
s3.upload_fileobj(f, "BUCKET_NAME", "OBJECT_NAME")
The code will use your environment variables for the access keys. Alternatively, you can use this syntax:
client = boto3.client(
's3',
aws_access_key_id=ACCESS_KEY,
aws_secret_access_key=SECRET_KEY,
aws_session_token=SESSION_TOKEN,
)