Lie Ryan wrote:
On 12/28/2009 5:42 PM, W. eWatson wrote:
You're right. Y. Works fine. The produces datetime.datetime(2009, 1, 2,
13, 1, 15).
If I now use
t2=datetime.datetime.strptime("2009/01/04 13:01:15","%Y/%m/%d %H:%M:%S")
I get tw as
datetime.datetime(2009, 1, 4, 13, 1, 15)
Then t2-t1 gives,
datetime.timedelta(2)
which is a 2 day difference--I guess. Strange.
what's strange about it? the difference between 2009/01/02 13:01:15 and
2009/01/04 13:01:15 is indeed 2 days... Can you elaborate what do you
mean by 'strange'?
Easily. In one case, it produces a one argument funcion, and the other
2, possibly even a year if that differs. How does one "unload" this
structure to get the seconds and days?
Changing
t2=datetime.datetime.strptime("2009/01/04 14:00:30","%Y/%m/%d %H:%M:%S")
and differencing gives me,
datetime.timedelta(2, 3555), which seems to indicate a 2 day and 3555
second difference. Interesting, but I think there must be another way to
do this. Maybe not.
to do... what?
To find the difference more clearly. Why not just return (0,2,3555)
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