打印两个日期之间的所有日期[重复]

问题描述：

例如：

from datetime import date

d1 = date(2008,8,15)
d2 = date(2008,9,15)

我正在寻找简单的代码来打印其间的所有日期：

2008,8,15  
2008,8,16  
2008,8,17  
...  
2008,9,14  
2008,9,15

谢谢

解决方案 1：

我想到了这个：

from datetime import date, timedelta

start_date = date(2008, 8, 15) 
end_date = date(2008, 9, 15)    # perhaps date.today()

delta = end_date - start_date   # returns timedelta

for i in range(delta.days + 1):
    day = start_date + timedelta(days=i)
    print(day)

输出：

2008-08-15
2008-08-16
...
2008-09-13
2008-09-14
2008-09-15

您的问题要求提供中间日期，但我相信您的意思是包括起点和终点，因此它们也包括在内。否则：

# To remove the end date, delete the "+ 1" 
# at the end of the range function:
for i in range(delta.days):

# To remove the start date, insert a 1 
# to the beginning of the range function:
for i in range(1, delta.days + 1):

解决方案 2：

使用列表推导：

from datetime import date, timedelta

d1 = date(2008,8,15)
d2 = date(2008,9,15)

# this will give you a list containing all of the dates
dd = [d1 + timedelta(days=x) for x in range((d2-d1).days + 1)]

for d in dd:
    print d

# you can't join dates, so if you want to use join, you need to
# cast to a string in the list comprehension:
ddd = [str(d1 + timedelta(days=x)) for x in range((d2-d1).days + 1)]
# now you can join
print "
".join(ddd)

解决方案 3：

本质上与 Gringo Suave 的答案相同，但带有生成器：

from datetime import datetime, timedelta


def datetime_range(start=None, end=None):
    span = end - start
    for i in xrange(span.days + 1):
        yield start + timedelta(days=i)

然后你可以按如下方式使用它：

In: list(datetime_range(start=datetime(2014, 1, 1), end=datetime(2014, 1, 5)))
Out: 
[datetime.datetime(2014, 1, 1, 0, 0),
 datetime.datetime(2014, 1, 2, 0, 0),
 datetime.datetime(2014, 1, 3, 0, 0),
 datetime.datetime(2014, 1, 4, 0, 0),
 datetime.datetime(2014, 1, 5, 0, 0)]

或者像这样：

In []: for date in datetime_range(start=datetime(2014, 1, 1), end=datetime(2014, 1, 5)):
   ...:     print date
   ...:     
2014-01-01 00:00:00
2014-01-02 00:00:00
2014-01-03 00:00:00
2014-01-04 00:00:00
2014-01-05 00:00:00

解决方案 4：

import datetime

d1 = datetime.date(2008,8,15)
d2 = datetime.date(2008,9,15)
diff = d2 - d1
for i in range(diff.days + 1):
    print (d1 + datetime.timedelta(i)).isoformat()

解决方案 5：

import datetime

begin = datetime.date(2008, 8, 15)
end = datetime.date(2008, 9, 15)

next_day = begin
while True:
    if next_day > end:
        break
    print next_day
    next_day += datetime.timedelta(days=1)