3 回答
TA贡献1963条经验 获得超6个赞
>>> from datetime import datetime, timedelta
>>> d = datetime.strptime('2013-05-27', '%Y-%m-%d') # Monday
>>> t = timedelta((12 - d.weekday()) % 7)
>>> d + t
datetime.datetime(2013, 6, 1, 0, 0)
>>> (d + t).strftime('%Y-%m-%d')
'2013-06-01'
我使用(12 - d.weekday()) % 7给定日期与下一个星期六之间的天数来计算增量,因为weekday介于0(星期一)至6(星期日)之间,所以星期六为5。但是:
5和12是相同的模7(是的,我们一周有7天:-))
所以
12 - d.weekday()
在6到125 - d.weekday()
之间,在5到-1之间因此,这使我无法处理否定情况(星期日为-1)。
这是任何工作日的非常简单的版本(无需检查):
>>> def get_next_weekday(startdate, weekday):
"""
@startdate: given date, in format '2013-05-25'
@weekday: week day as a integer, between 0 (Monday) to 6 (Sunday)
"""
d = datetime.strptime(startdate, '%Y-%m-%d')
t = timedelta((7 + weekday - d.weekday()) % 7)
return (d + t).strftime('%Y-%m-%d')
>>> get_next_weekday('2013-05-27', 5) # 5 = Saturday
'2013-06-01'
TA贡献1847条经验 获得超11个赞
我发现这个摆非常有用。只需一行
In [4]: pendulum.now().next(pendulum.SATURDAY).strftime('%Y-%m-%d')
Out[4]: '2019-04-27'
请参阅下面的更多细节:
In [1]: import pendulum
In [2]: pendulum.now()
Out[2]: DateTime(2019, 4, 24, 17, 28, 13, 776007, tzinfo=Timezone('America/Los_Angeles'))
In [3]: pendulum.now().next(pendulum.SATURDAY)
Out[3]: DateTime(2019, 4, 27, 0, 0, 0, tzinfo=Timezone('America/Los_Angeles'))
In [4]: pendulum.now().next(pendulum.SATURDAY).strftime('%Y-%m-%d')
Out[4]: '2019-04-27'
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