Use python to get an example of an N day or N month date around the current date

preface

Recently, due to my work, I found a time component of Python, which is very easy to share! (Forget the author's name, here first thank you), the following words do not say, to see a detailed introduction.

Sample code:

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

''' Gets around the current date N Days or N The date of the month '''

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):
 '''''
  Gets around the current date N The date of the month 
 if n>0,  Gets the current date before N The date of the month 
 if n<0,  After getting the current date N The date of the month 
 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)

conclusion