如何在不使用库的情况下在python中按自定义月份增加日期时间

Question

我需要增加日期时间值的月份

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

当月份为12时,它变为13并且引发错误"月份必须在1..12".(我预计这一年会增加)

我想使用timedelta,但它不需要一个月的论点.有relativedelta python包,但我不想只为此安装它.还有使用strtotime的解决方案.

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

我不想将datetime转换为str然后转换为time,然后转换为datetime; 因此,它仍然是一个图书馆

有没有人像使用timedelta一样有任何好的和简单的解决方案？

Answer 1

这是使用dateutilrelativedelta将日期添加到日期的简短方法.

from datetime import datetime
from dateutil.relativedelta import relativedelta

date_after_month = datetime.today()+ relativedelta(months=1)
print 'Today: ',datetime.today().strftime('%d/%m/%Y')
print 'After Month:', date_after_month.strftime('%d/%m/%Y')

输出:

今天:01/03/2013

月之后:01/04/2013

一句警告:relativedelta(months=1)并relativedelta(month=1)有不同的含义.传递month=1将原始日期中的月份替换为1月,而传递months=1将在原始日期添加一个月.

注意:这将需要python-dateutil.要安装它,您需要在Linux终端中运行.

sudo apt-get update && sudo apt-get install python-dateutil

说明:在python中添加月份值

Answer 2

编辑 - 根据您对日期的评论,如果下个月的天数较少,则需要向下舍入,这是一个解决方案:

import datetime
import calendar

def add_months(sourcedate, months):
    month = sourcedate.month - 1 + months
    year = sourcedate.year + month // 12
    month = month % 12 + 1
    day = min(sourcedate.day, calendar.monthrange(year,month)[1])
    return datetime.date(year, month, day)

此外,如果您不担心小时,分钟和秒钟,您可以使用date而不是datetime.如果您担心小时,分钟和秒钟,则需要修改我的代码以使用datetime并从源到结果复制小时,分钟和秒.

Answer 3

这是我的盐:

current = datetime.datetime(mydate.year, mydate.month, 1)
next_month = datetime.datetime(mydate.year + (mydate.month / 12), ((mydate.month % 12) + 1), 1)

快捷方便 :)

Answer 4

既然没有人提出任何解决方案,这就是我迄今为止的解决方法

year, month= divmod(mydate.month+1, 12)
if month == 0: 
      month = 12
      year = year -1
next_month = datetime.datetime(mydate.year + year, month, 1)

Answer 5

使用monthdelta包,它就像timedelta一样,但是对于日历月而不是几天/小时/等.

这是一个例子:

from monthdelta import MonthDelta

def prev_month(date):
    """Back one month and preserve day if possible"""
    return date + MonthDelta(-1)

与DIY方法相比:

def prev_month(date):
    """Back one month and preserve day if possible"""
   day_of_month = date.day
   if day_of_month != 1:
           date = date.replace(day=1)
   date -= datetime.timedelta(days=1)
   while True:
           try:
                   date = date.replace(day=day_of_month)
                   return date
           except ValueError:
                   day_of_month -= 1               

Answer 6

也许使用calendar.monthrange()添加当月的天数？

import calendar, datetime

def increment_month(when):
    days = calendar.monthrange(when.year, when.month)[1]
    return when + datetime.timedelta(days=days)

now = datetime.datetime.now()
print 'It is now %s' % now
print 'In a month, it will be %s' % increment_month(now)

Answer 7

这个如何？（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)

Answer 8

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)

如果您只想下个月的第一天：

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

Answer 9

要计算当前,上个月和下个月:

import datetime
this_month = datetime.date.today().month
last_month = datetime.date.today().month - 1 or 12
next_month = (datetime.date.today().month + 1) % 12 or 12

Answer 10

最简单的解决方案是在月底（我们总是知道每月至少有 28 天）并添加足够的天数以移至下个月：

>>> from datetime import datetime, timedelta
>>> today = datetime.today()
>>> today
datetime.datetime(2014, 4, 30, 11, 47, 27, 811253)
>>> (today.replace(day=28) + timedelta(days=10)).replace(day=today.day)
datetime.datetime(2014, 5, 30, 11, 47, 27, 811253)

也适用于年份之间：

>>> dec31
datetime.datetime(2015, 12, 31, 11, 47, 27, 811253)
>>> today = dec31
>>> (today.replace(day=28) + timedelta(days=10)).replace(day=today.day)
datetime.datetime(2016, 1, 31, 11, 47, 27, 811253)

请记住，不能保证下个月会是同一天，例如，从 1 月 31 日移动到 2 月 31 日时，它将失败：

>>> today
datetime.datetime(2016, 1, 31, 11, 47, 27, 811253)
>>> (today.replace(day=28) + timedelta(days=10)).replace(day=today.day)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: day is out of range for month

因此，如果您需要移至下个月的第一天，这是一个有效的解决方案，因为您始终知道下个月有第1天( .replace(day=1))。否则，要移至最后可用日期，您可能需要使用：

>>> today
datetime.datetime(2016, 1, 31, 11, 47, 27, 811253)
>>> next_month = (today.replace(day=28) + timedelta(days=10))
>>> import calendar
>>> next_month.replace(day=min(today.day, 
                               calendar.monthrange(next_month.year, next_month.month)[1]))
datetime.datetime(2016, 2, 29, 11, 47, 27, 811253)