使用python查找以秒为单位的时间差

Question

我需要用python找到秒的时差.我知道我可以得到这样的差异:

from datetime import datetime
now = datetime.now()
....
....
....
later = datetime.now()
difference = later-now

如何在总秒数上获得差异？

Answer 1

import time
now = time.time()
...
later = time.time()
difference = int(later - now)

Answer 2

total_seconds方法将返回差异,包括任何小数部分.

from datetime import datetime
now = datetime.now()
...
later = datetime.now()
difference = (later - now).total_seconds()

如果需要,可以通过int()将其转换为整数

Answer 3

将timedelta元组的项与足够的乘数相加应该会给您答案。 diff.days*24*60*60 + difference.seconds

from datetime import datetime
now = datetime.now()
...
later = datetime.now()
diff = later-now
diff_in_seconds = diff.days*24*60*60 + diff.seconds

变量'diff'是一个timedelta对象，是（天，秒，微秒）元组，如此处https://docs.python.org/2.4/lib/datetime-timedelta.html所述。所有其他单位（小时，分钟..）都转换为该格式。

>> diff = later- now
>> diff
datetime.timedelta(0, 8526, 689000)
>> diff_in_seconds = diff.days*24*60*60 + diff.seconds
>> diff_in_seconds
>> 8527

另一种看待它的方法是，如果您不是负数时差（因此是一个正时差），而不是现在较晚（现在是正的时差），则两者之间经过的时间仍然与前面的时间相同例

>> diff = earlier-now
>> diff
datetime.timedelta(-1, 77873, 311000)
>> diff_in_seconds = diff.days*24*60*60 + diff.seconds
>> diff_in_seconds
>> -8527

因此，即使我们确定工期少于1天，也必须考虑白天，因为这在负时差的情况下是很重要的。