获取从本周最后一个星期日到今天的日期列表

Question

我想要一个列表，其中包含一周中最后一个星期日到今天之间的所有日期，比如说今天是2019年11月5日，最后一个周日是2019年3月11日，我想要列出2019年11年3月3日到2019年5月11日的所有日期

我可以用下面的代码获得本周的最后一个星期天

from datetime import datetime, timedelta, date
today = date.today()
offset = (today.weekday() - 6) % 7
print("Offset", offset)
last_sunday = today - timedelta(days=offset)
print(last_sunday) 

我想要的输出是

['03/11/2019', '04/11/2019', '05/11/2019']

Answer 1

此处。希望这对你有帮助！

from datetime import datetime, timedelta, date
def daterange(date1, date2):
for n in range(int ((date2 - date1).days)+1):
yield date1 + timedelta(n)
today = date.today()
offset = (today.weekday() - 6) % 7
print("Offset", offset)
last_sunday = today - timedelta(days=offset)
print(last_sunday)
for dt in daterange(last_sunday, today):
print(dt.strftime("%Y-%m-%d"))

Answer 2

它的解决方案具有指定的输出格式。

from datetime import datetime, timdelta 
current_date = datetime.now()-timedelta(days=1)
datetime_list = [current_date-timedelta(days=offset) for offset in range((current_date.weekday()+1)%7+1)]
datetime_list.reverse()
datetime_list = [ '{0:02d}/{1:02d}/{2:}'.format(i.day,i.month,i.year) for i in datetime_list]
print(datetime_list)

Answer 3

from datetime import date, timedelta
sdate = date(2019,11,3)
today = date.today()
diff = today - sdate
for i in range(diff.days+1):
day = sdate + timedelta(days=i)
print(day)