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)))