tomorrow in yyyymmdd format

[Conrad]

Years ago, Nostradamus predicted that on Tue, 18 Jun 2002 20:52:25 -0500,
Gerhard Häring would write, saying:

> * Sean 'Shaleh' Perry <shalehperry at attbi.com> [2002-06-18 18:38 -0700]:
>> >> 
>> >> Sure there is :-) time.mktime() can take a "malformed" time tuple
>> >> and do something sensible with it. You can therefore get a time
>> >> tuple from gmtime, add one to the day position and call
>> >> time.mktime() on the result:
>> >> 
>> >> >>> x = (2002, 2, 29, 0, 31, 42, 2, 170, 0)
>> >> >>> time.gmtime(time.mktime(x))
>> >> (2002, 3, 1, 7, 31, 42, 4, 60, 0)
>> >> >>>
>> >> >>>
>> > Pure luck that this works. Don't complain if it stops workin in
>> > Python 2.7 and your code breaks ;-)
>> > 
>> > 
>> actually that is a function of the mktime() function in the C library
>> which the python module wraps.  Based on my reading of a man page or
>> two this behaviour seems guaranteed by POSIX.  Now, the assumption made
>> is the mktime() function on YOUR system follows POSIX.
> 
> Fine. I've just read the Python documentation which doesn't make this
> guarantee. I'll leave it to others to check if this works on win32,
> MacOS, as I'm using mxDateTime for this kind of tasks, anyway.
> 
> Gerhard


Hmm  - here's what I've done - a bit klutzy but it seems to work:

from time import *
one_day = 60*60*24                       # number of seconds in day
   # make a julian for tomorrow -
   # note: depending on what you want be careful around midnight!
tomorrow = mktime(localtime()) + one_day
print 'tomorrow is',localtime(tomorrow)
print 'today is',localtime()


tomorrow is (2002, 6, 27, 10, 42, 11, 3, 178, 1)
today is (2002, 6, 26, 10, 42, 11, 3, 178, 1)

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