3 回答
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TA贡献1776条经验 获得超12个赞
>>> import datetime
>>> a = datetime.datetime.now()
>>> b = datetime.datetime.now()
>>> c = b - a
datetime.timedelta(0, 8, 562000)
>>> divmod(c.days * 86400 + c.seconds, 60)
(0, 8) # 0 minutes, 8 seconds
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TA贡献1866条经验 获得超5个赞
使用datetime示例
>>> from datetime import datetime
>>> then = datetime(2012, 3, 5, 23, 8, 15) # Random date in the past
>>> now = datetime.now() # Now
>>> duration = now - then # For build-in functions
>>> duration_in_s = duration.total_seconds() # Total number of seconds between dates
持续时间多年
>>> years = divmod(duration_in_s, 31556926)[0] # Seconds in a year=31556926.
持续时间(天)
>>> days = duration.days # Build-in datetime function
>>> days = divmod(duration_in_s, 86400)[0] # Seconds in a day = 86400
持续时间(小时)
>>> hours = divmod(duration_in_s, 3600)[0] # Seconds in an hour = 3600
持续时间(分钟)
>>> minutes = divmod(duration_in_s, 60)[0] # Seconds in a minute = 60
持续时间(秒)
>>> seconds = duration.seconds # Build-in datetime function
>>> seconds = duration_in_s
持续时间,以微秒为单位
>>> microseconds = duration.microseconds # Build-in datetime function
两个日期之间的总持续时间
>>> days = divmod(duration_in_s, 86400) # Get days (without [0]!)
>>> hours = divmod(days[1], 3600) # Use remainder of days to calc hours
>>> minutes = divmod(hours[1], 60) # Use remainder of hours to calc minutes
>>> seconds = divmod(minutes[1], 1) # Use remainder of minutes to calc seconds
>>> print("Time between dates: %d days, %d hours, %d minutes and %d seconds" % (days[0], hours[0], minutes[0], seconds[0]))
或者干脆:
>>> print(now - then)
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