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

使用divmod:

now = int(time.time()) # epoch seconds
then = now - 90000 # some time in the past

d = divmod(now-then,86400)  # days
h = divmod(d[1],3600)  # hours
m = divmod(h[1],60)  # minutes
s = m[1]  # seconds

print '%d days, %d hours, %d minutes, %d seconds' % (d[0],h[0],m[0],s)

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”

我利用时差进行了连续集成测试，以检查和完善我的功能。如果有人需要，这里有一些简单的代码

from datetime import datetime

class TimeLogger:
    time_cursor = None

    def pin_time(self):
        global time_cursor
        time_cursor = datetime.now()

    def log(self, text=None) -> float:
        global time_cursor

        if not time_cursor:
            time_cursor = datetime.now()

        now = datetime.now()
        t_delta = now - time_cursor

        seconds = t_delta.total_seconds()

        result = str(now) + ' tl -----------> %.5f' % seconds
        if text:
            result += "   " + text
        print(result)

        self.pin_time()

        return seconds


time_logger = TimeLogger()

使用:

from .tests_time_logger import time_logger
class Tests(TestCase):
    def test_workflow(self):
    time_logger.pin_time()

    ... my functions here ...

    time_logger.log()

    ... other function(s) ...

    time_logger.log(text='Tests finished')

我在log输出中有类似的东西

2019-12-20 17:19:23.635297 tl -----------> 0.00007
2019-12-20 17:19:28.147656 tl -----------> 4.51234   Tests finished

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