Python的时间、日期和时间戳的处理办法

1.将字符串的时间转换为时间戳    方法:        a = "2013-10-10 23:40:00"        将其转换为时间数组        import time        timeArray = time.strptime(a, "%Y-%m-%d %H:%M:%S")转换为时间戳:timeStamp = int(time.mktime(timeArray))timeStamp == 1381419600

2.字符串格式更改如a = "2013-10-10 23:40:00",想改为 a = "2013/10/10 23:40:00"方法:先转换为时间数组,然后转换为其他格式timeArray = time.strptime(a, "%Y-%m-%d %H:%M:%S")otherStyleTime = time.strftime("%Y/%m/%d %H:%M:%S", timeArray)3.时间戳转换为指定格式日期:方法一:利用localtime()转换为时间数组,然后格式化为需要的格式,如timeStamp = 1381419600timeArray = time.localtime(timeStamp)otherStyleTime = time.strftime("%Y-%m-%d %H:%M:%S", timeArray)otherStyletime == "2013-10-10 23:40:00"方法二:import datetimetimeStamp = 1381419600dateArray = datetime.datetime.utcfromtimestamp(timeStamp)otherStyleTime = dateArray.strftime("%Y-%m-%d %H:%M:%S")otherStyletime == "2013-10-10 23:40:00"

注意：使用此方法时必须先设置好时区，否则有时差4.获取当前时间并转换为指定日期格式方法一:import time获得当前时间时间戳now = int(time.time())  ->这是时间戳转换为其他日期格式,如:"%Y-%m-%d %H:%M:%S"timeArray = time.localtime(timeStamp)otherStyleTime = time.strftime("%Y-%m-%d %H:%M:%S", timeArray)方法二:import datetime获得当前时间now = datetime.datetime.now()  ->这是时间数组格式转换为指定的格式:otherStyleTime = now.strftime("%Y-%m-%d %H:%M:%S")5.获得三天前的时间方法:import timeimport datetime先获得时间数组格式的日期threeDayAgo = (datetime.datetime.now() - datetime.timedelta(days = 3))转换为时间戳:timeStamp = int(time.mktime(threeDayAgo.timetuple()))转换为其他字符串格式:otherStyleTime = threeDayAgo.strftime("%Y-%m-%d %H:%M:%S")注:timedelta()的参数有:days,hours,seconds,microseconds6.给定时间戳,计算该时间的几天前时间:timeStamp = 1381419600先转换为datetimeimport datetimeimport timedateArray = datetime.datetime.utcfromtimestamp(timeStamp)threeDayAgo = dateArray - datetime.timedelta(days = 3)参考5,可以转换为其他的任意格式了7. 给定日期字符串，直接转换为datetime对象dateStr = '2013-10-10 23:40:00'datetimeObj = datetime.datetime.strptime(dateStr, "%Y-%m-%d %H:%M:%S")        注：将字符串日期转换为datetime后可以很高效的进行统计操作，因为转换为datetime后，           可以通过datetime.timedelta()方法来前后移动时间，效率很高，而且可读性很强。8.计算两个datetime之间的差距

       a = datetime.datetime(2014,12,4,1,59,59)

       b = datetime.datetime(2014,12,4,3,59,59)
       diffSeconds = (b-a).total_seconds()