在这个函数中，n可以是正数，也可以是负数。

def addmonth(d, n):
    n += 1
    dd = datetime.date(d.year + n/12, d.month + n%12, 1)-datetime.timedelta(1)
    return datetime.date(dd.year, dd.month, min(d.day, dd.day))

“Python -dateutil”(外部扩展)是一个很好的解决方案，但你可以使用内置的Python模块(datetime和datetime)来实现它。

我做了一个简短的代码，来解决它(处理年，月和日)

(运行:Python 3.8.2)

from datetime import datetime
from calendar import monthrange

# Time to increase (in months)
inc = 12

# Returns mod of the division for 12 (months)
month = ((datetime.now().month + inc) % 12) or 1

# Increase the division by 12 (months), if necessary (+ 12 months increase)
year = datetime.now().year + int((month + inc) / 12)

# (IF YOU DON'T NEED DAYS,CAN REMOVE THE BELOW CODE)
# Returns the same day in new month, or the maximum day of new month
day = min(datetime.now().day,monthrange(year, month)[1])

print("Year: {}, Month: {}, Day: {}".format(year, month, day))

嗯，这取决于你说的6个月后的日期。

使用自然月份: Inc = 6 年=年+(月+ inc - 1) // 12 月份=(月份+ inc - 1) % 12 + 1 用银行家的定义，6*30: 日期+=日期时间。Timedelta (6 * 30)

另一种解决方案:计算下个月的天数总和，并将结果添加到当前日期。

import calendar
import datetime

def date_from_now(months):
    today = datetime.datetime.today()

    month = today.month
    year = today.year
    sum_days = 0

    for i in range(int(months)):
        month += 1

        if month == 13:
            month = 1
            year += 1

        sum_days += calendar.monthrange(year, month)[1]

    return datetime.date.today() + datetime.timedelta(sum_days)

print(date_from_now(12)) # if to day is 2017-01-01, output: 2019-01-01 

按月开始计算:

from datetime import timedelta
from dateutil.relativedelta import relativedelta

end_date = start_date + relativedelta(months=delta_period) + timedelta(days=-delta_period)

修改了AddMonths()在Zope中使用和处理无效的天数:

def AddMonths(d,x):
    days_of_month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    newmonth = ((( d.month() - 1) + x ) % 12 ) + 1
    newyear  = d.year() + ((( d.month() - 1) + x ) // 12 ) 
    if d.day() > days_of_month[newmonth-1]:
      newday = days_of_month[newmonth-1]
    else:
      newday = d.day() 
    return DateTime( newyear, newmonth, newday)