Python: Run a code after return something-CodePudding

This is an ApiView from my django project:

class Receive_Payment(APIView):
    authentication_classes = (TokenAuthentication,)
    permission_classes = (IsAdminUser,)
    def get(self, request, cc_number):
        if Card.objects.filter(cc_num = cc_number).exists():
            client = Client.objects.get(client_id = (Card.objects.get(cc_num = cc_number).client))
            if not client.is_busy:
                client.is_busy = True
                client.save()
                resp = "successful"
            else:
                resp = "client_is_busy"
        else:
            resp = "fail"
        return Response({"response": resp})

As you see, if client.is_busy is not True, I'm making it True in here. But in this situation, I need to client.is_busy = False after 30 seconds. If I do it under client.save() code, it delays to response. How can I do it?

CodePudding user response：

Instead of using a boolean for determining if the client is busy, use a datetime and if it was set less than 30 seconds ago the client is busy. This then means you don't need to run any code after the request

from django.db import models
from django.utils.timezone import now
from datetime import timedelta


class Client(models.Model):
    last_updated = models.DateTimeField()

    @property
    def is_busy(self):
        # Can add a property to maintain the same is_busy boolean attribute
        return (now() - self.last_updated) < timedelta(seconds=30)

    def set_busy(self):
        self.last_updated = now()
        self.save()

CodePudding user response：

You should consider using timer objects here.

e.g.

import threading
def free_up_client(client):
   client.is_busy = false

timer = Theading.timer(30.0, free_up_client, [client])