如何在今年从Python开始打印 [英] How to Print next year from current year in Python

问题描述

如果使用最简单的代码在python中给出当前年份，则可以打印下一年，可能在一行中使用datetime模块。

How can I print the next year if the current year is given in python using the simplest code, possibly in one line using datetime module.

推荐答案

date和datetime对象都有一个年份属性，一个号码。只需添加1：

Both date and datetime objects have a year attribute, which is a number. Just add 1:

>>> from datetime import date
>>> print date.today().year + 1
2013

如果你有当年在一个变量中，只需直接添加1，不需要麻烦datetime模块：

If you have the current year in a variable, just add 1 directly, no need to bother with the datetime module:

>>> year = 2012
>>> print year + 1
2013

如果您的日期是字符串，只需选择代表年份的4位数字并将其传递给 int ：

If you have the date in a string, just select the 4 digits that represent the year and pass it to int:

>>> date = '2012-06-26'
>>> print int(date[:4]) + 1
2013

年算术非常简单，使其成为一个整数，只需添加1.它不会比这更简单。

Year arithmetic is exceedingly simple, make it an integer and just add 1. It doesn't get much simpler than that.

但是，如果您正在使用整个日期，而您需要同一天，但一年后，使用组件创建一个新的日期对象，年增加1：

If, however, you are working with a whole date, and you need the same date but one year later, use the components to create a new date object with the year incremented by one:

>>> today = date.today()
>>> print date(today.year + 1, today.month, today.day)
2013-06-26

，或者您可以使用 .replace 函数，该函数返回您指定的字段更改的副本：

or you can use the .replace function, which returns a copy with the field you specify changed:

>>> print today.replace(year=today.year + 1)
2013-06-26

请注意，当今天的今天是在2月29日的闰年，这可能会有点棘手。因此，绝对的，故障安全的正确方法是这样工作的：

Note that this can get a little tricky when today is February 29th in a leap year. The absolute, fail-safe correct way to work this one is thus:

def nextyear(dt):
   try:
       return dt.replace(year=dt.year+1)
   except ValueError:
       # February 29th in a leap year
       # Add 365 days instead to arrive at March 1st
       return dt + timedelta(days=365)

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