你说6个月是什么意思?

2009-02-13 + 6个月== 2009-08-13?还是2009-02-13 + 6*30天?

import mx.DateTime as dt

#6 Months
dt.now()+dt.RelativeDateTime(months=6)
#result is '2009-08-13 16:28:00.84'

#6*30 days
dt.now()+dt.RelativeDateTime(days=30*6)
#result is '2009-08-12 16:30:03.35'

更多关于mx的信息。DateTime

我发现这个解决方法很好。(使用python-dateutil扩展名)

from datetime import date
from dateutil.relativedelta import relativedelta

six_months = date.today() + relativedelta(months=+6)

这种方法的优势在于，它可以处理28天、30天、31天的问题。这在处理业务规则和场景(比如发票生成等)时非常有用。

$ date(2010,12,31)+relativedelta(months=+1)
  datetime.date(2011, 1, 31)

$ date(2010,12,31)+relativedelta(months=+2)
  datetime.date(2011, 2, 28)

下面是一个示例，它允许用户决定如何返回一个日期，其中一天大于一个月中的天数。

def add_months(date, months, endOfMonthBehaviour='RoundUp'):
    assert endOfMonthBehaviour in ['RoundDown', 'RoundIn', 'RoundOut', 'RoundUp'], \
        'Unknown end of month behaviour'
    year = date.year + (date.month + months - 1) / 12
    month = (date.month + months - 1) % 12 + 1
    day = date.day
    last = monthrange(year, month)[1]
    if day > last:
        if endOfMonthBehaviour == 'RoundDown' or \
            endOfMonthBehaviour == 'RoundOut' and months < 0 or \
            endOfMonthBehaviour == 'RoundIn' and months > 0:
            day = last
        elif endOfMonthBehaviour == 'RoundUp' or \
            endOfMonthBehaviour == 'RoundOut' and months > 0 or \
            endOfMonthBehaviour == 'RoundIn' and months < 0:
            # we don't need to worry about incrementing the year
            # because there will never be a day in December > 31
            month += 1
            day = 1
    return datetime.date(year, month, day)


>>> from calendar import monthrange
>>> import datetime
>>> add_months(datetime.datetime(2016, 1, 31), 1)
datetime.date(2016, 3, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2)
datetime.date(2015, 12, 1)
>>> add_months(datetime.datetime(2016, 1, 31), -2, 'RoundDown')
datetime.date(2015, 11, 30)

当我需要添加几个月或几年的时间，并且不想导入更多的库时，我就会这样做。

import datetime
__author__ = 'Daniel Margarido'


# Check if the int given year is a leap year
# return true if leap year or false otherwise
def is_leap_year(year):
    if (year % 4) == 0:
        if (year % 100) == 0:
            if (year % 400) == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False


THIRTY_DAYS_MONTHS = [4, 6, 9, 11]
THIRTYONE_DAYS_MONTHS = [1, 3, 5, 7, 8, 10, 12]

# Inputs -> month, year Booth integers
# Return the number of days of the given month
def get_month_days(month, year):
    if month in THIRTY_DAYS_MONTHS:   # April, June, September, November
        return 30
    elif month in THIRTYONE_DAYS_MONTHS:   # January, March, May, July, August, October, December
        return 31
    else:   # February
        if is_leap_year(year):
            return 29
        else:
            return 28

# Checks the month of the given date
# Selects the number of days it needs to add one month
# return the date with one month added
def add_month(date):
    current_month_days = get_month_days(date.month, date.year)
    next_month_days = get_month_days(date.month + 1, date.year)

    delta = datetime.timedelta(days=current_month_days)
    if date.day > next_month_days:
        delta = delta - datetime.timedelta(days=(date.day - next_month_days) - 1)

    return date + delta


def add_year(date):
    if is_leap_year(date.year):
        delta = datetime.timedelta(days=366)
    else:
        delta = datetime.timedelta(days=365)

    return date + delta


# Validates if the expected_value is equal to the given value
def test_equal(expected_value, value):
    if expected_value == value:
        print "Test Passed"
        return True

    print "Test Failed : " + str(expected_value) + " is not equal to " str(value)
    return False

# Test leap year
print "---------- Test leap year ----------"
test_equal(True, is_leap_year(2012))
test_equal(True, is_leap_year(2000))
test_equal(False, is_leap_year(1900))
test_equal(False, is_leap_year(2002))
test_equal(False, is_leap_year(2100))
test_equal(True, is_leap_year(2400))
test_equal(True, is_leap_year(2016))

# Test add month
print "---------- Test add month ----------"
test_equal(datetime.date(2016, 2, 1), add_month(datetime.date(2016, 1, 1)))
test_equal(datetime.date(2016, 6, 16), add_month(datetime.date(2016, 5, 16)))
test_equal(datetime.date(2016, 3, 15), add_month(datetime.date(2016, 2, 15)))
test_equal(datetime.date(2017, 1, 12), add_month(datetime.date(2016, 12, 12)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 31)))
test_equal(datetime.date(2015, 3, 1), add_month(datetime.date(2015, 1, 31)))
test_equal(datetime.date(2016, 3, 1), add_month(datetime.date(2016, 1, 30)))
test_equal(datetime.date(2016, 4, 30), add_month(datetime.date(2016, 3, 30)))
test_equal(datetime.date(2016, 5, 1), add_month(datetime.date(2016, 3, 31)))

# Test add year
print "---------- Test add year ----------"
test_equal(datetime.date(2016, 2, 2), add_year(datetime.date(2015, 2, 2)))
test_equal(datetime.date(2001, 2, 2), add_year(datetime.date(2000, 2, 2)))
test_equal(datetime.date(2100, 2, 2), add_year(datetime.date(2099, 2, 2)))
test_equal(datetime.date(2101, 2, 2), add_year(datetime.date(2100, 2, 2)))
test_equal(datetime.date(2401, 2, 2), add_year(datetime.date(2400, 2, 2)))

只需创建一个datetime.date()对象，调用add_month(date)来添加一个月，调用add_year(date)来添加一个年。

我是这样解决这个问题的:

import calendar
from datetime import datetime
moths2add = 6
now = datetime.now()
current_year = now.year
current_month = now.month
#count days in months you want to add using calendar module
days = sum(
  [calendar.monthrange(current_year, elem)[1] for elem in range(current_month, current_month + moths)]
    )
print now + days

我的实现基于taleinat的答案:

import datetime
import calendar

def add_months(orig_date, month_count = 1):
    while month_count > 12:
        month_count -= 12
        orig_date = add_months(orig_date, 12)
    new_year = orig_date.year
    new_month = orig_date.month + month_count
    # note: in datetime.date, months go from 1 to 12
    if new_month > 12:
        new_year += 1
        new_month -= 12

    last_day_of_month = calendar.monthrange(new_year, new_month)[1]
    new_day = min(orig_date.day, last_day_of_month)

    return orig_date.replace(year=new_year, month=new_month, day=new_day)

有了这个功能，你可以添加任意多的月份。

from datetime import date
dt = date(2021, 1, 31)

print(add_months(dt, 49))

返回 2025-02-28