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使用python时间处理方法

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在实际中遇到一个时间处理问题,需要将 Sep 06, 2014 19:30 (UTC 时间) 和 当前时间比较早晚,知道 此 2014-09-06 19:30 格式时间的运算。因此,在处理时,就想

先将sep 格式时间转化成 后面一个格式的时间样子,没有找到相关函数,于是自己简单的写了个程序,仅记于此,以备查询

代码如下:

# -*- coding: utf-8 -*- 
from datetime import date
from datetime import datetime
from datetime import timedelta
 
# # #如果是返回当前时间,可以简单的写成
# # time.localtime()
# # #这个返回UTC时间
# # time.gmtime()
# lt = time.localtime()
# tm = time.gmtime()
# ft = time.strftime('%Y-%m-%d %H-%M',lt)
# ft2 = time.strftime('%Y-%m-%d %H:%M',tm)
# print ft, ft2
# print '--------------------------------------------------------'
# now = datetime.datetime.now()
# now = now.replace(day = 1)
# print now
# print now.time()
#
# mytime = ['2014-09-06 20:19']
# #mytime2 = '2014-10-09 14:32'
# str = "".join(mytime)
# print str
# retime = datetime.strptime(str,'%Y-%m-%d %H:%M')
# print retime
# retime = retime +timedelta(hours = 8)
# print retime
#
# tdtime = datetime.now()
# print tdtime
# if retime <= tdtime - timedelta(days = 7):
# print "too early"
#
# #Sep 06, 2014 19:30
 
monthdic = {'Jan':'01', 'Feb':'02', 'Mar':'03', 'Apr':'04', 'May':'05', 'Jun':'06', 'Jul':'07', 'Aug':'08', 'Sep':'09', 'Oct':'10', 'Nov':'11', 'Dec':'12'}
def time_format(timestr):
 timestr = timestr.replace(',','')
 #print timestr
 timelist = timestr.split()
 #print timelist
 mon = "".join(timelist[0])
 #print mon
 timelist[0] = monthdic[mon]
 #print timelist
 mytime = "".join(timelist[2])+'-'+"".join(timelist[0])+'-'+"".join(timelist[1])+' '+ "".join(timelist[3])
 return mytime
if name == 'main':
 timestr = 'Sep 06, 2014 19:30'
 str = time_format(timestr)
 print str 
 mytime = datetime.strptime(str,'%Y-%m-%d %H:%M') 
 print mytime
 mytime = mytime +timedelta(hours = 8)
 print mytime 
 tdtime = datetime.now()
 print tdtime
 if mytime <= tdtime - timedelta(days = 7):
 print "too early"

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