Geek asks and answers

Celery Delay Callback

I use django-celery. I need to upload a large video file. I want to update my database when the file completes the download. Is there a way to add a callback that will call the django code and not another task when the task completes? My perfect code would look like this ...

from video.tasks import video_download
from video.models import Video

def my_callback(v):    
    v.status = "downloaded"
    v.save()

def download_http(request):
   v = Video.objects.latest().id #this is a string
   a = video_download.delay(v, my_callback)

If there is another way to update the object after completing the celery task, I will also be interested.

PS: I tried to switch to v = Video.objects.latest()instead v = Video.objects.latest().id, so I could just update the instance along the way, but the celery didn't like it because it was an object, not a string. Although it did not a.readyreturn any errors, every time I called , it returned False.

+3
source share
1

Django . :

def video_download(v):
     from video.models import Video
     v = Video.objects.get(pk=v)
     do_download(v)
     v.status = "downloaded"
     v.save()
+1

More articles:


All Articles