将epoch之后的时间戳转换为datetime.datetime

Question

我从epoch开始有以下时间戳:

Timestamp
1346114717972
1354087827000

如何将这些时间戳转换为某种特定的输出格式,例如mm/dd/yyyy hr:min:sec？

我试图将它们转换为datetime.datetime但失败了:

 >>> datetime.datetime.fromtimestamp(1346114717972)
 Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
 ValueError: timestamp out of range for platform time_t

我怎样才能做到这一点？

Answer 1

我会用这个time模块

>>> import time
>>> time.gmtime(1346114717972/1000.)
time.struct_time(tm_year=2012, tm_mon=8, tm_mday=28, tm_hour=0, tm_min=45, tm_sec=17, tm_wday=1, tm_yday=241, tm_isdst=0)

时间戳除以1000,因为您提供的标记是自纪元以来的毫秒数,而不是秒

然后使用strftime格式如此

>>> time.strftime('%m/%d/%Y %H:%M:%S',  time.gmtime(1346114717972/1000.))
'08/28/2012 00:45:17'

Answer 2

假设毫秒分辨率:

import datetime

s = '1346114717972'
fmt = "%Y-%m-%d %H:%M:%S"

# local time
t = datetime.datetime.fromtimestamp(float(s)/1000.)
print t.strftime(fmt) # prints 2012-08-28 02:45:17

# utc time
t_utc = datetime.datetime.utcfromtimestamp(float(s)/1000.)
print t_utc.strftime(fmt) # prints 2012-08-28 00:45:17

看看有关strftime()和strptime()行为的文档.

Answer 3

这是我见过的最简单的方法了

$ python
Python 2.7.5 (default, Nov  6 2016, 00:28:07) 
[GCC 4.8.5 20150623 (Red Hat 4.8.5-11)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import time
>>> print(time.strftime('%Y-%m-%dT%H:%M:%S %Z',time.localtime(time.time())))
2018-05-02T13:21:44 IST