weekend datetime objects

[Luke Skibinski Holt]

Is there a means of getting a datetime object as an offset from the
current date?
I need to get the datetime objects for the coming weekend, but if my
result is greater than the number of days in a month it fails rather
than wrapping into the next month. mx.DateTime and datetime.datetime
don't seem to support any such thing - has anybody else done such a
thing or could recommend a fix?

dt = datetime.now()
dt.replace(day=28, month=2)
wd = dt.weekday()
sat = dt.replace(day=dt.day+(5-wd))
sun = dt.replace(day=dt.day+(6-wd))

Traceback (most recent call last):
File "<stdin>", line 1, in ?
ValueError: day is out of range for month

Thanks,

Luke Skibinski Holt

[Malcolm Tredinnick]

Probably more of a question for comp.lang.python, but the answer is
pretty simple: import timedelta from datetime as well and then use

sat = dt + timedelta(days = 5 - dt.weekday())

This handles month and year wrapping without any problem.

Cheers,
Malcolm

[Russell Cloran]

Hi,

On Wed, 2006-02-22 at 19:12 -0800, Luke Skibinski Holt wrote:

> dt = datetime.now()
> dt.replace(day=28, month=2)
> wd = dt.weekday()
> sat = dt.replace(day=dt.day+(5-wd))
> sun = dt.replace(day=dt.day+(6-wd))

>>> from datetime import datetime, timedelta
>>> dt = datetime(2006, 2, 28)
>>> wd = dt.weekday()
>>> wd
1
>>> sat = dt + timedelta(5-wd)
>>> sat
datetime.datetime(2006, 3, 4, 0, 0)

HTH,

Russell
--
echo http://russell.rucus.net/spam/ |sed 's,t/.*,t,;P;s,.*//,,;s,\.,@,;'

[Luke Skibinski Holt]

Excellent - thanks for those - it's easy to get convoluted code when
you're thinking on an entirely different tack.

Luke Skibinski Holt

[Luke Skibinski Holt]

Despite being more appropriate for comp.lang.python, this is just an
example I came across for anyone that follows this thread:

# Adding to or Subtracting from a Date
# Use the rather nice datetime.timedelta objects

now = datetime.date(2003, 8, 6)
difference1 = datetime.timedelta(days=1)
difference2 = datetime.timedelta(weeks=-2)

print "One day in the future is:", now + difference1
#=> One day in the future is: 2003-08-07

print "Two weeks in the past is:", now + difference2
#=> Two weeks in the past is: 2003-07-23

print datetime.date(2003, 8, 6) - datetime.date(2000, 8, 6)
#=> 1095 days, 0:00:00

Taken directly from:
http://pleac.sourceforge.net/pleac_python/datesandtimes.html

Hope this helps someone.

Luke Skibinski Holt

[akaihola]

I have a related problem:

I like to calculate time differences with a custom SQL query to
optimize speed. The db cursor seems to return a DateTimeDelta object
(is that a psycopg class?) which I need to convert to a
datetime.timedelta to interoperate with other time objects in Django.

On the other hand, if I query DateTimeFields with custom SQL, Django
does automatically convert them to datetime.datetime objects.

Is there any particular reason why time differences returned by SQL
queries shouldn't be converted to datetime.timedelta objects by Django?

Here's what I have to do now (better ideas appreciated):

class Event(meta.Model):
starttime = meta.DateTimeField()
endtime = meta.DateTimeField()
def duration_minutes(self):
cursor = db.cursor()
cursor.execute("SELECT endtime-starttime FROM myapp_events
WHERE id = %s", [self.id])
try:
return cursor.fetchone()[0].minutes
# fetchone() returns e.g. (<DateTimeDelta object for
'02:54:00.00' at b756efc0>,)
except:
return 0.0
def duration(self):
from datetime import timedelta
return timedelta(minutes=self.duration_minutes())