I have been referring to the document https://docs.microsoft.com/en-us/azure/storage/blobs/storage-quickstart-blobs-python. I have not been able to find the proper APIs for
You should take a look at start_copy_from_url method in the SDK.
From the same link:
# Get the blob client with the source blob
source_blob = "<source-blob-url>"
copied_blob = blob_service_client.get_blob_client("<target-container>", '<blob-name>')
# start copy and check copy status
copy = copied_blob.start_copy_from_url(source_blob)
props = copied_blob.get_blob_properties()
print(props.copy.status)
Here's a full example for version 12.0.0 of the SDK:
from azure.storage.blob import BlobClient, BlobServiceClient
from azure.storage.blob import ResourceTypes, AccountSasPermissions
from azure.storage.blob import generate_account_sas
connection_string = '' # The connection string for the source container
account_key = '' # The account key for the source container
source_container_name = '' # Name of container which has blob to be copied
blob_name = '' # Name of the blob you want to copy
destination_container_name = '' # Name of container where blob will be copied
# Create client
client = BlobServiceClient.from_connection_string(connection_string)
# Create sas token for blob
sas_token = generate_account_sas(
account_name = client.account_name,
account_key = account_key
resource_types = ResourceTypes(object=True, container=True),
permission= AccountSasPermission(read=True,list=True),
start = datetime.now()
expiry = datetime.utcnow() + timedelta(hours=4) # Token valid for 4 hours
)
# Create blob client for source blob
source_blob = BlobClient(
client.url,
container_name = source_container_name,
blob_name = blob_name,
credential = sas_token
)
# Create new blob and start copy operation.
new_blob = client.get_blob_client(destination_container_name, blob_name)
new_blob.start_copy_from_url(source_blob.url)
See here for more information on how you can get the connection string and access key for the container.
This answer assumes that both containers are within the same subscription.