我需要增加一个datetime值的月份

next_month = datetime.datetime(mydate.year, mydate.month+1, 1)

当月份为12时,它变成13,并引发错误“month必须在1..12”。(我预计时间会增加)

我想使用timedelta,但它不带month参数。 有一个relativedelta python包,但我不想只为此安装它。 还有一个使用strtotime的解决方案。

time = strtotime(str(mydate));
next_month = date("Y-m-d", strtotime("+1 month", time));

我不想从datetime转换为str再转换为time,再转换为datetime;因此,它仍然是一个图书馆

有人有像使用timedelta一样好的简单的解决方案吗?


当前回答

好的,通过一些调整和使用timedelta,我们开始:

from datetime import datetime, timedelta


def inc_date(origin_date):
    day = origin_date.day
    month = origin_date.month
    year = origin_date.year
    if origin_date.month == 12:
        delta = datetime(year + 1, 1, day) - origin_date
    else:
        delta = datetime(year, month + 1, day) - origin_date
    return origin_date + delta

final_date = inc_date(datetime.today())
print final_date.date()

其他回答

我的解决方案非常简单,不需要任何额外的模块:

def addmonth(date):
    if date.day < 20:
        date2 = date+timedelta(32)
    else :
        date2 = date+timedelta(25)
    date2.replace(date2.year, date2.month, day)
    return date2

这个怎么样?(不需要任何额外的库)

from datetime import date, timedelta
from calendar import monthrange

today = date.today()
month_later = date(today.year, today.month, monthrange(today.year, today.month)[1]) + timedelta(1)

这是我想到的

from calendar  import monthrange

def same_day_months_after(start_date, months=1):
    target_year = start_date.year + ((start_date.month + months) / 12)
    target_month = (start_date.month + months) % 12
    num_days_target_month = monthrange(target_year, target_month)[1]
    return start_date.replace(year=target_year, month=target_month, 
        day=min(start_date.day, num_days_target_month))

与Dave Webb的解决方案的理想相似,但没有所有棘手的模运算:

import datetime, calendar

def increment_month(date):
    # Go to first of this month, and add 32 days to get to the next month
    next_month = date.replace(day=1) + datetime.timedelta(32)
    # Get the day of month that corresponds
    day = min(date.day, calendar.monthrange(next_month.year, next_month.month)[1])
    return next_month.replace(day=day)
from datetime import timedelta
try:
    next = (x.replace(day=1) + timedelta(days=31)).replace(day=x.day)
except ValueError:  # January 31 will return last day of February.
    next = (x + timedelta(days=31)).replace(day=1) - timedelta(days=1)

如果你只想要下个月的第一天:

next = (x.replace(day=1) + timedelta(days=31)).replace(day=1)