Print all day-dates between two dates [duplicate]
I came up with this:
from datetime import date, timedelta
start_date = date(2008, 8, 15)
end_date = date(2008, 9, 15) # perhaps date.now()
delta = end_date - start_date # returns timedelta
for i in range(delta.days + 1):
day = start_date + timedelta(days=i)
print(day)
The output:
2008-08-15
2008-08-16
...
2008-09-13
2008-09-14
2008-09-15
Your question asks for dates in-between but I believe you meant including the start and end points, so they are included. To remove the end date, delete the "+ 1" at the end of the range function. To remove the start date, insert a 1 argument to the beginning of the range function.
Using a list comprehension:
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)
Essentially the same as Gringo Suave's answer, but with a generator:
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)
Then you can use it as follows:
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)]
Or like this:
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()