How to update/insert query database automatic in django

Question

This is a digital library system. When borrow end_date has expired, borrow status will be updated automatically to 1 (End), table book column stock will be updated adding one stock, and automatically inserting new content value in table notification.

class Borrow(models.Model):
status = models.CharField(max_length=1, choices=[('0', 'Process'),('1', 'End')], default=0)
book = models.ForeignKey('models.Book', on_delete=models.CASCADE, related_name='borrow_book')
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
end_date = models.DateTimeField()
    
class Book(models.Model):
stock = models.IntegerField(blank=True, null=True)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)

class Notification(models.Model):
content = models.CharField(max_length=255, blank=True, null=True)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)

How to make an automatic update query in table borrow column status, table book column stock, and insert in table notification when end_date > DateTime.now() ?

Answer 1

I would make a management command and then create a Cron / Scheduled Task to run it Hourly.
There are packages like django-crontab and django-cron that could also do this (I assume?), but I've never used either and have stuck with the ol' Default Linux Crontab

Note: I left out creating Notification

Management Command

from django.core.management.base import BaseCommand
from django.core.management import call_command

class Command(BaseCommand):
    help = 'Ran Hourly With a Cron'

    def handle(self, *args, **kwargs):
        from app.models import Barrow
        from datetime import datetime

        # Fetch List
        l = Borrow.objects.filter(status='0', end_date__lte=datetime.now())

        # Update stock
        #   F() means you don't need to fetch the value to update it!
        from django.db.models import F
        for i in l:
            b = i.book
            b.stock += F('stock') + 1
            b.save()

        # update List
        l.update(status='1')

This would probably be a better way to update the count, especially if there's multiple barrows for a single book.

from django.db.models import Count
for i in l.values('book').annotate(count=Count('pk')):
    # i = { 'book': {book__pk}, 'count': total_int }
    b = Book.objects.get(pk=i['book'])
    b.stock += F('stock') + i['count']
    b.save()

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But this might not be the best idea because a book could be late to be returned and then your count is off.