打印两个date之间的所有date
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打印之间的所有date：
2008,8,15 2008,8,16 2008,8,17 ... 2008,9,14 2008,9,15
谢谢
我想出了这个：
from datetime import date, timedelta d1 = date(2008, 8, 15) # start date d2 = date(2008, 9, 15) # end date delta = d2 - d1 # timedelta for i in range(delta.days + 1): print(d1 + timedelta(days=i))
输出：
2008-08-15 2008-08-16 ... 2008-09-13 2008-09-14 2008-09-15
你的问题要求在中间的date，但我相信你的意思包括开始和结束点，所以他们包括在内。 要删除结束date，请删除for循环结尾处的+1。 要删除开始date，请在范围函数的开头添加1。
使用列表理解：
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 "\n".join(ddd)
与Gringo Suave的答案基本相同，但是有一个生成器：
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)
那么你可以使用它如下：
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
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()
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)
