下面是一个稍微不同的答案，基于S.Lott的答案，给出了两个日期开始和结束之间的日期列表。在下面的例子中，从2017年初到今天。

start = datetime.datetime(2017,1,1)
end = datetime.datetime.today()
daterange = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]

如果有两个日期，你需要范围试试

from dateutil import rrule, parser
date1 = '1995-01-01'
date2 = '1995-02-28'
datesx = list(rrule.rrule(rrule.DAILY, dtstart=parser.parse(date1), until=parser.parse(date2)))

下面是一个稍微不同的答案，基于S.Lott的答案，给出了两个日期开始和结束之间的日期列表。在下面的例子中，从2017年初到今天。

start = datetime.datetime(2017,1,1)
end = datetime.datetime.today()
daterange = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]

from datetime import datetime , timedelta, timezone


start_date = '2022_01_25'
end_date = '2022_01_30'

start = datetime.strptime(start_date, "%Y_%m_%d")
print(type(start))
end =  datetime.strptime(end_date, "%Y_%m_%d")
##pDate = str(pDate).replace('-', '_')
number_of_days = (end - start).days

print("number_of_days: ", number_of_days)

##
date_list = []
for day in range(number_of_days):
    a_date = (start + timedelta(days = day)).astimezone(timezone.utc)
    a_date = a_date.strftime('%Y-%m-%d')
    date_list.append(a_date)

print(date_list)

我想用一个简单(不完整)的日期范围实现来发表我的意见:

from datetime import date, timedelta, datetime

class DateRange:
    def __init__(self, start, end, step=timedelta(1)):
        self.start = start
        self.end = end
        self.step = step

    def __iter__(self):
        start = self.start
        step = self.step
        end = self.end

        n = int((end - start) / step)
        d = start

        for _ in range(n):
            yield d
            d += step

    def __contains__(self, value):
        return (
            (self.start <= value < self.end) and 
            ((value - self.start) % self.step == timedelta(0))
        )

你可以写一个生成器函数，返回从今天开始的日期对象:

import datetime

def date_generator():
  from_date = datetime.datetime.today()
  while True:
    yield from_date
    from_date = from_date - datetime.timedelta(days=1)

这个生成器返回从今天开始的日期，一次返回一天。以下是前3次约会的方法:

>>> import itertools
>>> dates = itertools.islice(date_generator(), 3)
>>> list(dates)
[datetime.datetime(2009, 6, 14, 19, 12, 21, 703890), datetime.datetime(2009, 6, 13, 19, 12, 21, 703890), datetime.datetime(2009, 6, 12, 19, 12, 21, 703890)]

与循环或列表推导相比，这种方法的优点是可以返回任意多次。

Edit

使用生成器表达式代替函数的更紧凑的版本:

date_generator = (datetime.datetime.today() - datetime.timedelta(days=i) for i in itertools.count())

用法:

>>> dates = itertools.islice(date_generator, 3)
>>> list(dates)
[datetime.datetime(2009, 6, 15, 1, 32, 37, 286765), datetime.datetime(2009, 6, 14, 1, 32, 37, 286836), datetime.datetime(2009, 6, 13, 1, 32, 37, 286859)]