Re: Datetime objects

[Rama]

I am a newcomer to python - so not too sure if this method is the correct one.

 

Noticing this,

 

>>> dir(diff)
['__abs__', '__add__', '__class__', '__delattr__', '__div__', '__doc__', '__eq__
', '__floordiv__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__
', '__le__', '__lt__', '__mul__', '__ne__', '__neg__', '__new__', '__nonzero__',
 '__pos__', '__radd__', '__rdiv__', '__reduce__', '__reduce_ex__', '__repr__', '
__rfloordiv__', '__rmul__', '__rsub__', '__setattr__', '__str__', '__sub__', 'da
ys', 'max', 'microseconds', 'min', 'resolution', 'seconds']

 

it seems

 

>>> diff.seconds
52662

 

should give you what you want. Converting the seconds to hours, minutes, seconds should be easy.

 

Of course, looking at help(diff), I get the impression that in the generic case you should check if diff.days == 0 to decide if you should be usingthat as well.

 

thank,

Rama

 

On 2 Aug 2006 05:16:08 -0700, Lad <[EMAIL PROTECTED]> wrote:

Sybren Stuvel wrote:
> Lad enlightened us with:
> > How can I find days and minutes difference between two datetime
> > objects?
> > For example If I  have
> > b=datetime.datetime(2006, 8, 2, 8, 57, 28, 687000)
> > a=datetime.datetime(2006, 8, 1, 18, 19, 45, 765000)
>
> diff = b - a

Ok, I tried

>>> diff=b-a
>>> diff
datetime.timedelta(0, 52662, 922000)
>>> diff.min
datetime.timedelta(-999999999)


which is not good for me.

So I tried to use toordinal like this
diff=b.toordinal()-a.toordinal()

but I get
diff=1

Why?
where do I make a mistake?
Thank you for help

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