如何在SQLAlchemy中仅比较日期时间的日期？

Question

我在SQLAlchemy有一个查询：

weekly_schedule = WeeklyHour.query.filter(
    and_(
        WeeklyHour.start_time.in_(week_dates),
        WeeklyHour.end_time.in_(week_dates),
        WeeklyHour.employee == employee
        )
    ).all()

我正在这样获取本周的日期：

today = datetime.datetime.today()
week_dates = [today + datetime.timedelta(days=i) for i in xrange(0 - today.weekday(), 7 - today.weekday())]

这将返回datetimes当前星期的列表。 像这样：

[datetime.date(2014, 4, 7), datetime.date(2014, 4, 8), datetime.date(2014, 4, 9), datetime.date(2014, 4, 10), datetime.date(2014, 4, 11), datetime.date(2014, 4, 12), datetime.date(2014, 4, 13)]

问题是， WeeklyHour是一个datetime ，当员工开始和结束的一天。 并且week_dates是设置为now()的特定时间。 我永远不会得到结果，因为datetime的time不匹配。 我真正想要比较的是datetime的date部分。 我当前的解决方案（有效但很愚蠢）是存储datetime单独dates 。

Answer 1

而不是查询一周中的所有日期，而是以天为限制。

today = date.today()
start_of_week = today - timedelta(days=today.weekday())
start_of_following_week = start_of_week + timedelta(days=7)

weekly_schedule = WeeklyHour.query.filter(
    and_(
        WeeklyHour.start_time >= start_of_week,  # On or after Monday
        WeeklyHour.end_time < start_of_following_week, # Before next Monday
        WeeklyHour.employee == employee
        )
    ).all()