利用python获取当前日期前后N天或N月日期的方法示例
发布时间:2020-12-17 08:12:40 所属栏目:Python 来源:网络整理
导读:前言 最近因为工作原因,发现一个Python的时间组件,很好用分享出来!(忘记作者名字了,在这里先感谢了),下面话不多说,来一起看看详细的介绍吧。 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,loc
前言 最近因为工作原因,发现一个Python的时间组件,很好用分享出来!(忘记作者名字了,在这里先感谢了),下面话不多说,来一起看看详细的介绍吧。 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,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,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,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,d) = getyearandmonth(n) arr = (y,d) if (int(day) < int(d)): arr = (y,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 以上就是这篇文章的全部内容了,希望本文的内容对大家的学习或者工作能带来一定的帮助,如果有疑问大家可以留言交流,谢谢大家对编程小技巧的支持 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |