利用python獲取當前日期前后N天或N月日期的方法示例

更新時間：2017年07月30日 08:24:41 投稿：daisy

最近在工作中遇到一個需求，查找資料發(fā)現(xiàn)了一個很好的時間組件，所以下面這篇文章主要給大家介紹了關于利用python獲取當前日期前后N天或N月日期的方法示例，需要的朋友們可以參考借鑒，下面來一起看看吧。

前言

最近因為工作原因，發(fā)現(xiàn)一個Python的時間組件，很好用分享出來?。ㄍ涀髡呙至?，在這里先感謝了），下面話不多說，來一起看看詳細的介紹吧。

示例代碼：

# -*- coding: utf-8 -*-

'''獲取當前日期前后N天或N月的日期'''

from time import strftime, localtime
from datetime import timedelta, date
import calendar

year = strftime("%Y", localtime())
mon = strftime("%m", localtime())
day = strftime("%d", localtime())
hour = strftime("%H", localtime())
min = strftime("%M", localtime())
sec = strftime("%S", localtime())

def today():
 '''''
 get today,date format="YYYY-MM-DD"
 '''''
 return date.today()


def todaystr():
 '''
 get date string, date format="YYYYMMDD"
 '''
 return year + mon + day


def datetime():
 '''''
 get datetime,format="YYYY-MM-DD HH:MM:SS"
 '''
 return strftime("%Y-%m-%d %H:%M:%S", localtime())


def datetimestr():
 '''''
 get datetime string
 date format="YYYYMMDDHHMMSS"
 '''
 return year + mon + day + hour + min + sec


def get_day_of_day(n=0):
 '''''
 if n>=0,date is larger than today
 if n<0,date is less than today
 date format = "YYYY-MM-DD"
 '''
 if (n < 0):
  n = abs(n)
  return date.today() - timedelta(days=n)
 else:
  return date.today() + timedelta(days=n)


def get_days_of_month(year, mon):
 '''''
 get days of month
 '''
 return calendar.monthrange(year, mon)[1]


def get_firstday_of_month(year, mon):
 '''''
 get the first day of month
 date format = "YYYY-MM-DD"
 '''
 days = "01"
 if (int(mon) < 10):
  mon = "0" + str(int(mon))
 arr = (year, mon, days)
 return "-".join("%s" % i for i in arr)


def get_lastday_of_month(year, mon):
 '''''
 get the last day of month
 date format = "YYYY-MM-DD"
 '''
 days = calendar.monthrange(year, mon)[1]
 mon = addzero(mon)
 arr = (year, mon, days)
 return "-".join("%s" % i for i in arr)


def get_firstday_month(n=0):
 '''''
 get the first day of month from today
 n is how many months
 '''
 (y, m, d) = getyearandmonth(n)
 d = "01"
 arr = (y, m, d)
 return "-".join("%s" % i for i in arr)


def get_lastday_month(n=0):
 '''''
 get the last day of month from today
 n is how many months
 '''
 return "-".join("%s" % i for i in getyearandmonth(n))


def getyearandmonth(n=0):
 '''''
 get the year,month,days from today
 befor or after n months
 '''
 thisyear = int(year)
 thismon = int(mon)
 totalmon = thismon + n
 if (n >= 0):
  if (totalmon <= 12):
   days = str(get_days_of_month(thisyear, totalmon))
   totalmon = addzero(totalmon)
   return (year, totalmon, days)
  else:
   i = totalmon / 12
   j = totalmon % 12
   if (j == 0):
    i -= 1
    j = 12
   thisyear += i
   days = str(get_days_of_month(thisyear, j))
   j = addzero(j)
   return (str(thisyear), str(j), days)
 else:
  if ((totalmon > 0) and (totalmon < 12)):
   days = str(get_days_of_month(thisyear, totalmon))
   totalmon = addzero(totalmon)
   return (year, totalmon, days)
  else:
   i = totalmon / 12
   j = totalmon % 12
   if (j == 0):
    i -= 1
    j = 12
   thisyear += i
   days = str(get_days_of_month(thisyear, j))
   j = addzero(j)
   return (str(thisyear), str(j), days)


def addzero(n):
 '''''
 add 0 before 0-9
 return 01-09
 '''
 nabs = abs(int(n))
 if (nabs < 10):
  return "0" + str(nabs)
 else:
  return nabs


def get_today_month(n=0):
 '''''
 獲取當前日期前后N月的日期
 if n>0, 獲取當前日期前N月的日期
 if n<0, 獲取當前日期后N月的日期
 date format = "YYYY-MM-DD"
 '''
 (y, m, d) = getyearandmonth(n)
 arr = (y, m, d)
 if (int(day) < int(d)):
  arr = (y, m, day)
 return "-".join("%s" % i for i in arr)


if __name__ == "__main__":
 print today()
 print todaystr()
 print datetime()
 print datetimestr()
 print get_day_of_day(20)
 print get_day_of_day(-3)
 print get_today_month(-3)
 print get_today_month(3)
 print get_today_month(19)

總結(jié)

以上就是這篇文章的全部內(nèi)容了，希望本文的內(nèi)容對大家的學習或者工作能帶來一定的幫助，如果有疑問大家可以留言交流，謝謝大家對腳本之家的支持

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