时间:2021-07-01 10:21:17 帮助过:83人阅读
from datetime import datetime, timedelta
weekdays = ['Monday','Tuesday','Wednesday','Thursday',
'Friday','Saturday','Sunday']
def get_previous_byday(dayname, start_date=None):
if start_date is None:
start_date = datetime.today()
day_num = start_date.weekday()
day_num_target = weekdays.index(dayname)
days_ago = (7 + day_num - day_num_target) % 7
if days_ago == 0:
days_ago = 7
target_date = start_date - timedelta(days = days_ago)
return target_date
print( datetime.today() )
print( get_previous_byday('Monday') )
print( get_previous_byday('Monday', datetime(2016, 8, 28)) )第二种方法,用dateutil模块
from datetime import datetime from dateutil.relativedelta import relativedelta from dateutil.rrule import * d = datetime.now() print(d) print(d + relativedelta(weekday=FR)) print(d + relativedelta(weekday=FR(-1)))