One way to upload a video in the background and receive a response without waiting for the upload to complete is to use asynchronous programming. You can use the celery
library along with a task queue to handle the video uploads asynchronously.
First, you need to create a Celery task to handle the video upload. This task should take in the necessary data, such as the file path and any other metadata, and perform the upload asynchronously. Once the upload is complete, the task can return a response to the API view indicating that the upload was successful.
Here is an example of what the Celery task might look like:
from celery import shared_task
from django.core.files.storage import default_storage
@shared_task
def upload_video(file_path, metadata):
# Perform the video upload asynchronously
with open(file_path, 'rb') as f:
default_storage.save(metadata['filename'], f)
return {'status': 'success'}
Then, in your API view, you can call this task using the delay()
method, which queues the task to be executed asynchronously:
class VideoUploadView(APIView):
serializer_class = VideoSerializer
def post(self, request, *args, **kwargs):
serializer = self.serializer_class(data=request.data)
if serializer.is_valid():
# Get the file path and metadata from the serializer
file_path = serializer.validated_data['file_path']
metadata = serializer.validated_data['metadata']
# Queue the upload task to be executed asynchronously
upload_video.delay(file_path, metadata)
# Return a response indicating that the upload was successful
return Response({'status': 'success'})
# If the serializer is not valid, return an error response
return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)
In this example, the view simply queues the upload task and returns a response indicating that the upload was successful. The actual upload process takes place asynchronously in the Celery task, allowing the view to return quickly without waiting for the upload to complete.
Asked: 2021-06-29 11:00:00 +0000
Seen: 7 times
Last updated: Aug 31 '21