Python实现获取当前日期前后N天或N月日期的方法示例
本文分享一个基于Python标准库的时间处理示例,涵盖获取当前日期、时间字符串及前后N天/N月日期的功能。通过结合datetime、time和calendar模块,实现灵活的日期偏移计算与格式化输出,适用于日常开发中的时间戳转换需求。
前言
近期因工作需要,发现一个Python时间处理组件十分实用,特此分享。虽已遗忘原作者姓名,但仍在此表示感谢。以下将详细介绍具体实现内容。
示例代码:
# -*- coding: utf-8 -*-
'''获取当前日期前后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):
'''''
获取当前日期前后N月的日期
if n>0, 获取当前日期前N月的日期
if n<0, 获取当前日期后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)
总结
以上内容为本文全部介绍,期望对读者的学习或工作有所助益。若有任何疑问,欢迎留言交流,感谢支持。
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