Python根据时间获取周数与下周日期范围代码实例

import time
import datetime
# 获取今天是第几周
print(time.strftime('%W'))
# 获取当前是周几(0-6,0代表周一)
today=datetime.datetime.now().weekday()
# 获取指定日期属于当年的第几周
week=datetime.datetime.strptime('20190825','%Y%m%d').strftime('%W')

import datetime,calendar,pandas as pd
def get_N_day():
 # 获取当前日期
 today1 = datetime.date.today()
 today2 = datetime.date.today
 # 指定时间周期
 oneday = datetime.timedelta(days=1)
 oneweek = datetime.timedelta(days=7)
 # 获取周一和周日的判断条件（周一为0，周日为6）
 m1 = calendar.MONDAY
 m2 = calendar.SUNDAY
 # 获取下周周一的时间
 if today1.weekday() != m1:
  while today1.weekday() != m1:
   today1 += oneday
 else:
  today1+=oneweek
 # 获取下周周日的时间
 if today2.weekday() != m2:
  today2 += oneweek
  while today2.weekday() != m2:
   today2 += oneday
 else:
  today2 += oneweek
 # 转换时间形式
 nextMonday = today1.strftime('%Y%m%d')
 nextSunday = today2.strftime('%Y%m%d')
 # 获取下周七天的时间列表
 date_list = [d.strftime("%Y-%m-%d") for d in pd.date_range(nextMonday, nextSunday, freq="D")]
 # 转换呈现形式
 list_date=[]
 for i in date_list:
  one=i.split('-')
  time=['月','日']
  time.insert(0,one[1])
  time.insert(2,one[2])
  time1=''.join(time)
  list_date.append(time1)
 return list_date
print(get_N_day())