在Django的午夜减少模型值 [英] Decrementing model value at midnight in Django

问题描述

我想在午夜每天将值days_till_study减1。

I want to decrement the value days_till_study with 1 everyday at midnight.

我目前的方法不是很优雅。有更好的方法吗？

My current approach is not very elegant. Is there a better way to do this?

from django.db import models
from django.utils import timezone
    
class Card(models.Model):
    question = models.CharField(max_length=100)
    answer = models.TextField()
    date = models.DateTimeField(default=timezone.now)
    creator = models.ForeignKey(User, on_delete=models.CASCADE)
    decks = models.ManyToManyField(Deck)
    days_till_study = models.IntegerField(default=1)

    def __str__(self):
        return self.question

    def decrement_days_till_study(self):
        if days_till_study < 1:
            x = str(datetime.datetime.now())
            if x[x[11:26]] == '00:00:00.000000':
                days_till_study += 1

感谢阅读。

推荐答案

即时计算结果

如果您可以控制存储在数据库中的内容，最好将两天都存储在数据库中并根据它计算差异：

Calculating the result on the fly

If you have control over what to store in the database, it's better to store both days in the database and calculate the difference based on it:

class Card(models.Model):
    created_at = models.DateTimeField()
    study_at = models.DateTimeField()

    @property
    def days_till_study(self):
        return (self.study_at - self.created_at).days

通过Celery计划

如果您仍然需要按计划执行某项操作（例如更新在午夜> days_till_study ），您可以尝试 django-cron 库或其他基于Celery的解决方案。

Scheduling via Celery

If you still need to do something by a schedule (like updating days_till_study at midnight) you can try django-cron library or another Celery-based solution.

此外，您还可以创建管理命令，并通过系统 cron 。

Also, you can create a management command and run it by a schedule via the system cron.

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