这就是我如何获得两个datetime之间经过的小时数。datetime对象:

before = datetime.datetime.now()
after  = datetime.datetime.now()
hours  = math.floor(((after - before).seconds) / 3600)

用其他方法得到日期之间的差异;

import dateutil.parser
import datetime
last_sent_date = "" # date string
timeDifference = current_date - dateutil.parser.parse(last_sent_date)
time_difference_in_minutes = (int(timeDifference.days) * 24 * 60) + int((timeDifference.seconds) / 60)

得到最小值的输出。

谢谢

基于@Attaque的精彩回答，我提出了一个更短的简化版本的日期时差计算器:

seconds_mapping = {
    'y': 31536000,
    'm': 2628002.88, # this is approximate, 365 / 12; use with caution
    'w': 604800,
    'd': 86400,
    'h': 3600,
    'min': 60,
    's': 1,
    'mil': 0.001,
}

def get_duration(d1, d2, interval, with_reminder=False):
    if with_reminder:
        return divmod((d2 - d1).total_seconds(), seconds_mapping[interval])
    else:
        return (d2 - d1).total_seconds() / seconds_mapping[interval]

我改变了它，以避免声明重复的函数，删除了漂亮的打印默认间隔，并增加了对毫秒、周和ISO月的支持(简单地说，月只是一个近似，基于每个月等于365/12的假设)。

生产:

d1 = datetime(2011, 3, 1, 1, 1, 1, 1000)
d2 = datetime(2011, 4, 1, 1, 1, 1, 2500)

print(get_duration(d1, d2, 'y', True))      # => (0.0, 2678400.0015)
print(get_duration(d1, d2, 'm', True))      # => (1.0, 50397.12149999989)
print(get_duration(d1, d2, 'w', True))      # => (4.0, 259200.00149999978)
print(get_duration(d1, d2, 'd', True))      # => (31.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'h', True))      # => (744.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'min', True))    # => (44640.0, 0.0014999997802078724)
print(get_duration(d1, d2, 's', True))      # => (2678400.0, 0.0014999997802078724)
print(get_duration(d1, d2, 'mil', True))    # => (2678400001.0, 0.0004999997244524721)

print(get_duration(d1, d2, 'y', False))     # => 0.08493150689687975
print(get_duration(d1, d2, 'm', False))     # => 1.019176965856293
print(get_duration(d1, d2, 'w', False))     # => 4.428571431051587
print(get_duration(d1, d2, 'd', False))     # => 31.00000001736111
print(get_duration(d1, d2, 'h', False))     # => 744.0000004166666
print(get_duration(d1, d2, 'min', False))   # => 44640.000024999994
print(get_duration(d1, d2, 's', False))     # => 2678400.0015
print(get_duration(d1, d2, 'mil', False))   # => 2678400001.4999995

你可能会发现这个快速的片段在不那么长的时间间隔里很有用:

    from datetime import datetime as dttm
    time_ago = dttm(2017, 3, 1, 1, 1, 1, 1348)
    delta = dttm.now() - time_ago
    days = delta.days # can be converted into years which complicates a bit…
    hours, minutes, seconds = map(int, delta.__format__('').split('.')[0].split(' ')[-1].split(':'))

在Python v.3.8.6上测试

要查找天数:timedelta有一个'days'属性。你可以简单地查询一下。

>>>from datetime import datetime, timedelta
>>>d1 = datetime(2015, 9, 12, 13, 9, 45)
>>>d2 = datetime(2015, 8, 29, 21, 10, 12)
>>>d3 = d1- d2
>>>print d3
13 days, 15:59:33
>>>print d3.days
13

只要用一个减去另一个。你会得到一个timedelta对象。

>>> import datetime
>>> d1 = datetime.datetime.now()
>>> d2 = datetime.datetime.now() # after a 5-second or so pause
>>> d2 - d1
datetime.timedelta(0, 5, 203000)
>>> dd = d2 - d1
>>> print (dd.days) # get days
>>> print (dd.seconds) # get seconds
>>> print (dd.microseconds) # get microseconds
>>> print (int(round(dd.total_seconds()/60, 0))) # get minutes