>>> import datetime
>>> first_time = datetime.datetime.now()
>>> later_time = datetime.datetime.now()
>>> difference = later_time - first_time
datetime.timedelta(0, 8, 562000)
>>> seconds_in_day = 24 * 60 * 60
>>> divmod(difference.days * seconds_in_day + difference.seconds, 60)
(0, 8)      # 0 minutes, 8 seconds

从第一个时间差中减去后面的时间= later_time - first_time创建一个只保存时间差的datetime对象。 在上面的例子中，它是0分钟，8秒和562000微秒。

这里有一个很容易概括或转化为函数的答案，它合理紧凑，易于遵循。

ts_start=datetime(2020, 12, 1, 3, 9, 45)
ts_end=datetime.now()
ts_diff=ts_end-ts_start
secs=ts_diff.total_seconds()
days,secs=divmod(secs,secs_per_day:=60*60*24)
hrs,secs=divmod(secs,secs_per_hr:=60*60)
mins,secs=divmod(secs,secs_per_min:=60)
secs=round(secs, 2)
answer='Duration={} days, {} hrs, {} mins and {} secs'.format(int(days),int(hrs),int(mins),secs)
print(answer)

它给出的答案是“持续时间=270天10小时32分42.13秒”

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)

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

这是我使用mktime的方法。

from datetime import datetime, timedelta
from time import mktime

yesterday = datetime.now() - timedelta(days=1)
today = datetime.now()

difference_in_seconds = abs(mktime(yesterday.timetuple()) - mktime(today.timetuple()))
difference_in_minutes = difference_in_seconds / 60

这可能会帮助一些人，用这个方法找到过期与否其计算天数。这是dt。秒和dt。微秒也可用

from datetime import datetime
# updated_at = "2022-10-20T07:18:56.950563"
def is_expired(updated_at):
    expires_in = 7 #days
    datetime_format = '%Y-%m-%dT%H:%M:%S.%f'
    time_difference = datetime.now() - datetime.strptime(updated_at, datetime_format)

    return True if time_difference.days > expires_in else False