python 时间处理之月份加减问题

目录

• python时间处理月份加减
• python日期相减（秒、小时、天、月、年）

python时间处理月份加减

第三方模块 ：

python-dateutil

安装方式：

pip install python-dateutil

实例代码：

import datetime
from dateutil.relativedelta importjs relativedelta

datetime_now = datetime.datetime.now()
datetime_three_month_ago = datetime_now - relativedelta(months=3)
print datetime_three_month_ago

python日期相减（秒、小时、天、月、年）

python代码

编程

import datetime
today=datetime.datetime.now()
datetime1=today-datetime.timedelta(seconds=10)#减10秒
datetime2=today-datetime.timedelta(minutes=10)#减10分钟
datetime3=today-datetime.timedelta(hours=1)#减1小时
datetime4=today-datetimphpe.timedelta(days=7)#减1天
datetime5=today-datetime.timedelta(weeks=1)#减1周
datetime6=today.strftime('%Y-%m-%d')#将时间格式化为字符串
print('today:',today)
print('second:',datetime1)
print('minute:',datetime2)
print('hour:',datetime3)
print('day:',datetime4)
print('week:',datetime5)
print('时间格式化为字符串:',datetime6)
 
 
import dateutil.relativedelta
datetime7= datetjavascriptime.datetime.strptime(datetime6, '%Y-%m-%d')#将字符串格式为时间
datetime8 = datetime7 - dateutil.relativedelta.relativedelta(seconds=10)#减10秒
datetime9 = datetime7 - dateutil.relativedelta.relativjavascriptedelta(minutes=10)#减10分钟
datetime10 = datetime7 - dateutil.relativedelta.relativedelta(hours=1)#减减1小时
datetime11 = datetime7 - dateutil.relativedelta.relativedelta(days=1)#减1天
datetime12 = datetime7 - dat开发者_开发入门eutil.relativedelta.relativedelta(months=1)#减一个月
datetime13 = datetime7 - dateutil.relativedelta.relativedelta(years=1)#减一年
print('字符串格式为时间:',datetime7)
print('second2:',datetime8)
print('minute2:',datetime9)
print('hour2:',datetime10)
print('day2:',datetime11)
print('month2:',datetime12)
print('year:',datetime13)

运行结果

以上为个人经验，希望能给大家一个参考，也希望大家多多支持我们。