在Python中查找时间

Question

找到时间直到某个日期的最佳方法是什么？我想知道几年，几个月，几天和几小时。

我希望有人有一个很好的功能。我想做类似这样的事情：此评论已于2个月和3天前发布，或者此评论已发布1年5个月前。

Answer 1

datetime模块，datetime和timedelta对象，它会给你几天和几秒钟。

In [5]: datetime.datetime(2009, 10, 19) - datetime.datetime.now() 
Out[5]: datetime.timedelta(2, 5274, 16000) 

In [6]: td = datetime.datetime(2009, 10, 19) - datetime.datetime.now() 

In [7]: td.days 
Out[7]: 2 

In [8]: td.seconds 
Out[8]: 5262 

Answer 2

您应该使用dateutil .relativedelta。

from dateutil.relativedelta import relativedelta 
import datetime 
today = datetime.date.today() 
rd = relativedelta(today, datetime.date(2001,1,1)) 
print "comment created %(years)d years, %(months)d months, %(days)d days ago" % rd.__dict__ 

Answer 3

我正在寻找更像这样的东西......这花了一些努力去寻找。

import datetime 

SECOND = 1 
MINUTE = 60 * SECOND 
HOUR = 60 * MINUTE 
DAY = 24 * HOUR 
MONTH = 30 * DAY 

def get_relative_time(dt):  
    now = datetime.datetime.now() 
    delta_time = dt - now 

    delta = delta_time.days * DAY + delta_time.seconds 
    minutes = delta/MINUTE 
    hours = delta/HOUR 
    days = delta/DAY 

    if delta < 0: 
     return "already happened" 

    if delta < 1 * MINUTE:  
     if delta == 1: 
      return "one second to go" 
     else: 
      return str(delta) + " seconds to go" 


    if delta < 2 * MINUTE:  
     return "a minute ago" 


    if delta < 45 * MINUTE:  
     return str(minutes) + " minutes to go" 

    if delta < 90 * MINUTE:  
     return "an hour ago" 

    if delta < 24 * HOUR: 
     return str(hours) + " hours to go" 

    if delta < 48 * HOUR:  
     return "yesterday" 

    if delta < 30 * DAY:  
     return str(days) + " days to go" 


    if delta < 12 * MONTH:  
     months = delta/MONTH 
     if months <= 1: 
      return "one month to go" 
     else: 
      return str(months) + " months to go" 
    else:  
     years = days/365.0 
     if years <= 1: 
      return "one year to go" 
     else: 
      return str(years) + " years to go" 

Answer 4

让我们asume你有一个名为ETA变量未来的日期时间：

(eta - datetime.datetime.now()).total_seconds() 

日期时间差异导致timedelta对象，这恰好实现了一个名为total_seconds方法。这是它:)

Answer 5

可能是你想要的东西是这样的：

import datetime 

today = datetime.date.today() 
futdate = datetime.date(2016, 8, 10) 

now  = datetime.datetime.now() 
mnight = now.replace(hour=0, minute=0, second=0, microsecond=0) 
seconds = (mnight - now).seconds 
days = (futdate - today).days 
hms  = str(datetime.timedelta(seconds=seconds)) 

print ("%d days %s" % (days, hms))