alw*_*btc 1 python time date subtraction
我有2个日期时间对象:
a
datetime.datetime(2013,2,11,15,35)
b
datetime.datetime(2013,2,11,18,55)
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差异是3:20
我怎么能在python中获得这个?
我做:
(b-a).seconds/3600
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但我得到3而不是3:20
手动结构可以接受吗?如果你减去两个日期时间,你将得到一个timedelta对象.然后从中获取值:
>>> diff = b - a
>>> diff
datetime.timedelta(0, 12000)
>>> hrs = diff.seconds / 3600
>>> hrs
3
>>> mins = diff.seconds % 3600 / 60
>>> mins
20
>>> "%s:%s" % (hrs, mins)
'3:20'
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