我对Tony Diep的答案的修改，可能稍微更优雅(当然，Python 2，匹配问题和原始答案的日期，对于Python 3，必要时修改，至少包括/ to //):

def add_months(date, months):
    month = date.month + months - 1
    year = date.year + (month / 12)
    month = (month % 12) + 1
    day = date.day
    while (day > 0):
        try:
            new_date = date.replace(year=year, month=month, day=day)
            break
        except:
            day = day - 1    
    return new_date

根据“业务需求”解释添加月份，即日期映射到月底之后，应该映射到月底，而不是下一个月。

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

这是我想到的。它移动了正确的月份和年份，但忽略了天数(这是我的情况所需要的)。

import datetime

month_dt = 4
today = datetime.date.today()
y,m = today.year, today.month
m += month_dt-1
year_dt = m//12
new_month = m%12
new_date = datetime.date(y+year_dt, new_month+1, 1)

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)

我经常需要一个月的最后一天来保持上个月的最后一天。为了解决这个问题，我在计算前加一天，然后在返回前再减去它。

from datetime import date, timedelta

# it's a lot faster with a constant day
DAY = timedelta(1)

def add_month(a_date, months):
    "Add months to date and retain last day in month."
    next_day = a_date + DAY
    # calculate new year and month
    m_sum = next_day.month + months - 1
    y = next_day.year + m_sum // 12
    m = m_sum % 12 + 1
    try:
        return date(y, m, next_day.day) - DAY
    except ValueError:
        # on fail return last day in month
        # can't fail on december so I don't bother changing the year
        return date(y, m + 1, 1) - DAY

假设你的datetime变量叫做date:

date=datetime.datetime(year=date.year+int((date.month+6)/12),
                       month=(date.month+6)%13 + (1 if (date.month + 
                       months>12) else 0), day=date.day)