PyQt4的QDate类有一个addmonths函数。

>>>from PyQt4.QtCore import QDate  
>>>dt = QDate(2009,12,31)  
>>>required = dt.addMonths(6) 

>>>required
PyQt4.QtCore.QDate(2010, 6, 30)

>>>required.toPyDate()
datetime.date(2010, 6, 30)

我们可能应该使用dateutil。relativedelta

然而，出于学术兴趣，我只想补充一点，在我发现它之前，我打算用这个:

亿: 今天代表。年度+ (K .今日month + 6) / 12 / (K . today . 12 month + 5%) + 1, K .今日day)。 except: 今天代表。年+ (K.today.month+6)//12 (K.today.month+6)%12+1, 1) -时间轴(天= 1)

它看起来很简单，但仍然可以捕捉到所有的问题，如29、30、31

它也适用于- 6 MTHS通过执行-timedelta

别被k弄糊涂了，今天它只是我程序中的一个变量

Dateutil包实现了这样的功能。但要知道，这将是天真的，因为其他人已经指出。

从这个答案中，请参见parsedatetime。下面是代码示例。更多细节:使用许多自然语言-> YYYY-MM-DD转换示例进行单元测试，以及明显的parsedatetime转换挑战/bug。

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import time, calendar
from datetime import date

# from https://github.com/bear/parsedatetime
import parsedatetime as pdt

def print_todays_date():
    todays_day_of_week = calendar.day_name[date.today().weekday()]
    print "today's date = " + todays_day_of_week + ', ' + \
                              time.strftime('%Y-%m-%d')

def convert_date(natural_language_date):
    cal = pdt.Calendar()
    (struct_time_date, success) = cal.parse(natural_language_date)
    if success:
        formal_date = time.strftime('%Y-%m-%d', struct_time_date)
    else:
        formal_date = '(conversion failed)'
    print '{0:12s} -> {1:10s}'.format(natural_language_date, formal_date)

print_todays_date()
convert_date('6 months')

上面的代码从MacOSX机器生成以下内容:

$ ./parsedatetime_simple.py 
today's date = Wednesday, 2015-05-13
6 months     -> 2015-11-13
$ 

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 

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)