我有一个餐厅和运营时间的模型,具有外键关系 . 例如,餐厅海景咖啡馆,其每周工作日的营业时间为
周五 - 上午10点 - 下午6:30
周四 - 上午10点 - 下午9:45
周三 - 上午10点 - 下午9:45
周二 - 上午10点 - 下午9:45
星期一 - 上午10点 - 下午9:45
周日 - 上午10点 - 晚上8点
我只能根据这样的开启和关闭时间进行检查
models.py
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)
@property
def open_or_closed(self):
operating_time = OperatingTime.objects.all()
opening_time = operating_time.opening_time
closing_time = operating_time.closing_time
current_time = datetime.now().time()
weekday = operating_time.day_of_week
if opening_time < current_time < closing_time:
open_or_closed = True
else:
open_or_closed = False
return open_or_closed
@property
def open_or_closed(self):
operating_time = OperatingTime.objects.all()
opening_time = operating_time.opening_time
closing_time = operating_time.closing_time
current_time = datetime.now().time()
weekday = operating_time.day_of_week
if opening_time < current_time < closing_time:
open_or_closed = True
else:
open_or_closed = False
return self.open_or_closed
如果所有餐厅的餐厅每天都开放或关闭,我该怎么办?
1 回答
没有测试但它应该工作 .