Question

我有两个模型用于餐厅，另一个用于操作时间。操作时间与餐厅有外键关系，因为一周内不同日子的操作时间可能不同。我想表明餐厅在当前时间和当天是开放还是关闭。在views.py中对此进行编码或为此创建模板标记会更好吗？因为惯例说视图应该很薄。

用于餐厅和营业时间的

模型

class Restaurant(models.Model):
    owner = models.ForeignKey(User)
    name = models.CharField(max_length=150, db_index=True)
    address = models.CharField(max_length=100)


class OperatingTime(models.Model):
    MONDAY = 1
    TUESDAY = 2
    WEDNESDAY = 3
    THURSDAY = 4
    FRIDAY = 5
    SATURDAY = 6
    SUNDAY = 7

    DAY_IN_A_WEEK = (
        (MONDAY, 'Monday'),
        (TUESDAY, 'Tuesday'),
        (WEDNESDAY, 'Wednesday'),
        (THURSDAY, 'Thursday'),
        (FRIDAY, 'Friday'),
        (SATURDAY, 'Saturday'),
        (SUNDAY, 'Sunday'),
        )
    # HOURS = [(i, i) for i in range(1, 25)]
    restaurant = models.ForeignKey(Restaurant,related_name="operating_time")
    opening_time = models.TimeField()
    closing_time = models.TimeField()
    day_of_week = models.IntegerField(choices=DAY_IN_A_WEEK)

    def __str__(self):
        return '{} ---- {}'.format(self.opening_time, self.closing_time)

views.py

def home(request):
    restaurant = Restaurant.objects.all()
    print('restaurant',restaurant)
    operating_time = OperatingTime.objects.all()
    print('operating time',operating_time)
    for operating_time in operating_time: # Tried to find if restaurant is opened or closed based on opening time & closing time in current time and day for each restaurant
        opening = operating_time.opening_time
        closing = operating_time.closing_time
        print('opening',opening)
    current_time = datetime.now()
    current_time = current_time.time()
    if current_time < closing or opening< current_time:
        print('opening')
    else:
        print('closed')
    return render(request, 'restaurant/homepage.html', {'restaurant':restaurant})

我怎么能找到这个？在视图上编码或创建模板标记更好吗？

Answer 1

我现在在我的工作地点与django工作了大约一个月，并且被教导不要在观点中加入任何逻辑。该视图应仅用于传递数据库中的数据或将数据发布到数据库。逻辑应该在模板标签中完成。

在当前时间和日期显示餐厅是否开放

1 个答案: