I want to create a work schedule; I will have to input some names
(in alphabetical order) and the days they can't be working because they
have a license, are in vacation or are doing special services. Then,
using that information, the program would assign, from a Monday
specified onwards, from two weeks to two weeks for a period of 3 months
(12 weeks), one person to the morning and two to
the afternoon, following the order.However, if the person is not available in any day of that two
weeks (input given earlier); it is not allocated to that two weeks of
work. Instead, it is still the first of the list (it will be assigned
in the future to the next two weeks when there's nothing to impede the
work).
Then the program will output the schedule to a file or the screen
(preferrably a file). I've included an example of only one period. The
impediment list is optional, but highly recommended if you can do it.If anyone would want to program it for me, I'd be pleased. However, if you just want to help me, I'll post the code and the problem:
day = int(raw_input("day: "))
month = int(raw_input("month: "))
year = int(raw_input("year: "))
for week in range(5):
if day in calendar.monthcalendar(ano, mes)[week]:
startweek = week
def definsemana():
monthcalendar = calendar.monthcalendar(year, month)
t1s = monthcalendar[startweek]
if t1s[-1] == 0:
startweek = 0
if month < 12:
month += 1
for i in range(7):
if t1s[i] == 0:
t1s[i] = calendar.monthcalendar(year, month)[startweek][i]
startweek = 1
else:
month = 2
year += 1
for day in t1s:
print day, '\t',
This SHOULD get the day specified and output the week following it, however it seems that it's outputting the first week of the month, regardless of the input date (try 26 December 2005 as an example). Where's the problem?
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