Tastypie: Is there a way to invalidate Cache after resource Update?

Question

I have a Django (1.8.17) app, and I am using Tastypie (0.13.3) as REST API Framework. I have some very simple resources, and I use SimpleCache to cache them for 15 minutes.

from tastypie.resources import ModelResource
from my_app.models import MyModel
from tastypie.cache import SimpleCache


class MyModelResource(ModelResource):
    cache = SimpleCache(timeout=900)
    class Meta:
        queryset = MyModel.objects.all()

The problem: When I update the Resource by a PUT request, the resource gets updated in DataBase, but doesn't get invalidated from cache, so I continue to get the old data for 15 minutes, which is inconvenient, I would like that when the resource get updated, it will be fetched from DB and cached again. Is there someone who faced the same issue ?

Possible duplicate of [Why is Django returning stale cache data?](https://stackoverflow.com/questions/37849236/why-is-django-returning-stale-cache-data) — e4c5, May 30 '17 at 11:33
It is the packages are different but the core problem and the solution are the same — e4c5, May 30 '17 at 11:42
@e4c5 No the core Problem is not the same, and the solution is not really useful. I digged inside Tastypie and Django Caching and found an efficient and straightforward solution ! — Yahya Yahyaoui, May 30 '17 at 14:22

score 1 · Accepted Answer · answered May 30 '17 at 14:39

After a long search without finding any thing, I got an idea ! Override the PUT method by just deleting the object from cache each time it gets updated, and this is the natural way that should happen from the beginning. I found that Tastypie's SimpleCache is using Django's Core Cache (at least in this case; think settings), so here it is how I solved the problem:

from tastypie.resources import ModelResource
from my_app.models import MyModel
from tastypie.cache import SimpleCache
from django.core.cache import cache


class MyModelResource(ModelResource):
    cache = SimpleCache(timeout=900)
    class Meta:
        queryset = MyModel.objects.all()

    def put_detail(self, request, **kwargs):
        # Build the key
        key = "{0}:{1}:detail:pk={2}".format(kwargs['api_name'], kwargs['resource_name'], kwargs['pk'])
        cache.delete(key)
        return super(MyModelResource, self).put_detail(request, **kwargs)

Tastypie: Is there a way to invalidate Cache after resource Update?

1 Answers1