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python?時間處理之月份加減問題_python

作者:平頭哥(AdgerZhou) ? 更新時間: 2022-12-26 編程語言

python時間處理月份加減

第三方模塊 :

python-dateutil

安裝方式:

pip install python-dateutil

實例代碼:

import datetime
from dateutil.relativedelta import relativedelta
?
datetime_now = datetime.datetime.now()
datetime_three_month_ago = datetime_now - relativedelta(months=3)
print datetime_three_month_ago

python日期相減(秒、小時、天、月、年)

python代碼

import datetime
today=datetime.datetime.now()
datetime1=today-datetime.timedelta(seconds=10)#減10秒
datetime2=today-datetime.timedelta(minutes=10)#減10分鐘
datetime3=today-datetime.timedelta(hours=1)#減1小時
datetime4=today-datetime.timedelta(days=7)#減1天
datetime5=today-datetime.timedelta(weeks=1)#減1周
datetime6=today.strftime('%Y-%m-%d')#將時間格式化為字符串
print('today:',today)
print('second:',datetime1)
print('minute:',datetime2)
print('hour:',datetime3)
print('day:',datetime4)
print('week:',datetime5)
print('時間格式化為字符串:',datetime6)
 
 
import dateutil.relativedelta
datetime7= datetime.datetime.strptime(datetime6, '%Y-%m-%d')#將字符串格式為時間
datetime8 = datetime7 - dateutil.relativedelta.relativedelta(seconds=10)#減10秒
datetime9 = datetime7 - dateutil.relativedelta.relativedelta(minutes=10)#減10分鐘
datetime10 = datetime7 - dateutil.relativedelta.relativedelta(hours=1)#減減1小時
datetime11 = datetime7 - dateutil.relativedelta.relativedelta(days=1)#減1天
datetime12 = datetime7 - dateutil.relativedelta.relativedelta(months=1)#減一個月
datetime13 = datetime7 - dateutil.relativedelta.relativedelta(years=1)#減一年
print('字符串格式為時間:',datetime7)
print('second2:',datetime8)
print('minute2:',datetime9)
print('hour2:',datetime10)
print('day2:',datetime11)
print('month2:',datetime12)
print('year:',datetime13)

運行結(jié)果

原文鏈接:https://pingtouge.blog.csdn.net/article/details/79444094

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