实施“开放时间”的任何现有解决方案在Django [英] Any existing solution to implement "opening hours" in Django

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本文介绍了实施“开放时间”的任何现有解决方案在Django的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在为一个想要改变每个不同商店的开放时间
的客户的网站。 Django是否存在此类问题的解决方案?

I'm making a website for a client who wants to be able to change then opening hours for each of his different stores. Is there an existing solution for this type of problem with Django ?

推荐答案

你是什么意思?似乎很简单根据您的平日订单进行调整。如果你喜欢,添加验证。但是,人们应该足够聪明,不需要验证这种东西。

What do you mean? Seems pretty simple. Adjust according to your weekday order. And if you like, add validation. But people should be smart enough to not need validation for that sort of stuff.

WEEKDAYS = [
  (1, _("Monday")),
  (2, _("Tuesday")),
  (3, _("Wednesday")),
  (4, _("Thursday")),
  (5, _("Friday")),
  (6, _("Saturday")),
  (7, _("Sunday")),
]

class OpeningHours(models.Model):
    store = models.ForeignKey("StoreModel")
    weekday_from = models.IntegerField(choices=WEEKDAYS, unique=True)
    weekday_to = models.IntegerField(choices=WEEKDAYS)
    from_hour = models.IntegerField(choices=range(1,25))
    to_hour = models.IntegerField(choices=range(1,25))

    def get_weekday_from_display(self):
        return WEEKDAYS[self.weekday_from]

    def get_weekday_to_display(self):
        return WEEKDAYS[self.weekday_to]

class SpecialDays(models.Model):
    holiday_date = models.DateField()
    closed = models.BooleanField(default=True)
    from_hour = models.IntegerField(choices=range(1,25), null=True, blank=True)
    to_hour = models.IntegerField(choices=range(1,25), null=True, blank=True)

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