datetime.datetime - 日期字段超出范围

Question

I have found very useful datetime.datetime object when dealing with dates, however I have situation now where datime.datetime isn't working for me.During execution of the program, day field is dynamically calculated and here's the problem:

>>> datetime.datetime(2013, 2, 29, 10, 15)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: day is out of range for month

OK, February doesn't have 29 days, but would be great if datetime could figure that out and return this object

datetime.datetime(2013, 3, 1, 10, 15)

What's the best way to solve this situation ? So, I'm looking for a general solution, when day argument is bigger than number of days month could have.

Answer 1

From the Zen of Python: Explicit is better than implicit . When you make an error like trying to create an invalid date, you need to explicitly handle that situation.

How you handle that exception is entirely up to your application. You could inform the end user of the error, or you could try and shift the days into the next month, or cap the day to the last legal day in the current month. All would be valid options, depending on your use cases .

The following code would shift the 'surplus' days into a next month. So 2013-02-30 would become 2013-03-02 instead.

import calendar
import datetime

try:
    dt = datetime.datetime(year, month, day, hour, minute)
except ValueError:
    # Oops, invalid date. Assume we can fix this by shifting this to the next month instead
    _, monthdays = calendar.monthrange(year, month)
    if monthdays < day:
        surplus = day - monthdays
        dt = datetime.datetime(year, month, monthdays, hour, minute) + datetime.timedelta(days=surplus)

Answer 2

While there is much to be said about using try...except in this situation, if you really only need the month + daysOffset you can do this:

d = datetime.datetime(targetYear,targetMonth,1,hour,min,sec)
d = d + datetime.timedelta(days=targetDayOfMonth-1)

Basically, set the day of the month to 1, which is always in the month, then add timedelta to return the appropriate date in a current or future month.

d = datetime.datetime(2013, 2, 1, 10, 15) # day of the month is 1
# since the target day is the 29th and that is 28 days after the first
# subtract 1 before creating the timedelta.
d = d + datetime.timedelta(days=28) 
print d
# datetime.datetime(2013, 3, 1, 10, 15)

Answer 3

使用下个月的第一天然后减去一天以避免使用日历

datetime.datetime(targetYear, targetMonth+1, 1) + dt.timedelta(days = -1)