Python datetime subtraction - wrong results?

I must be doing something wrong here, any ideas?

>开发者_JAVA技巧;>> (datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,16)).seconds
2098

It should be getting more than that many seconds.

timedelta.seconds gives you the seconds field of the timedelta. But it also has a days field (and a milliseconds field).

So you would want something like

delta = datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,16)
delta.seconds + delta.days*86400

It's actually returning a timedelta which has a day field also i.e.

c.seconds = 2098

but

c.days = 1

I wonder that there's still no total_seconds example. It works for both python2 and python3. And it's the easiest way to get timedelta total seconds count.

In your case

In [5]: (datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,1
   ...: 6)).seconds
Out[5]: 2098

In [6]: (datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,1
   ...: 6)).total_seconds()
Out[6]: 88498.0

Another example:

In [1]: from datetime import datetime
In [2]: d1 = datetime(2018,1,1)
In [3]: d2 = datetime(2018,1,3)
In [4]: td = d2 - d1
In [5]: td.seconds
Out[5]: 0
In [6]: td.days
Out[6]: 2
In [7]: td.total_seconds()
Out[7]: 172800.0

timedelta.seconds isn't the total number of seconds, it's the remainder in seconds after days have been accounted for. Using your example:

>>> import datetime
>>> (datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,16))
datetime.timedelta(1, 2098)

That datetime.timedelta(1, 2098) means that your timedelta is 1 day, plus 2098 seconds.

What you want is something like:

>>> delta = (datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,16))
>>> (delta.days * 86400) + delta.seconds
88498

datetime.datetime(2008,11,7,10,5,14)-datetime.datetime(2008,11,6,9,30,16) returns a datetime.timedelta object which has a days attribute. The difference that you are calculating is actually 1 day and 2098 seconds.