问题：如何从gmtime（）的时间+日期输出获取自纪元以来的秒数？
你如何做相反gmtime()，你把时间+日期，并获得秒数？
我有类似的字符串'Jul 9, 2009 @ 20:02:58 UTC'，并且我想获取从纪元到2009年7月9日之间的秒数。
我已经尝试过，time.strftime但是我不知道如何正确使用它，或者它是否是正确的命令。
How do you do reverse gmtime(), where you put the time + date and get the number of seconds?
I have strings like 'Jul 9, 2009 @ 20:02:58 UTC', and I want to get back the number of seconds between the epoch and July 9, 2009.
I have tried time.strftime but I don’t know how to use it properly, or if it is the correct command to use.
回答 0
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如果想扭转time.gmtime()，那就想calendar.timegm()。
>>> calendar.timegm(time.gmtime())
1293581619.0
您可以使用将字符串转换为时间元组time.strptime()，该时间元组返回一个时间元组，您可以将其传递给calendar.timegm()：
>>> import calendar
>>> import time
>>> calendar.timegm(time.strptime('Jul 9, 2009 @ 20:02:58 UTC', '%b %d, %Y @ %H:%M:%S UTC'))
1247169778
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If you want to reverse time.gmtime(), you want calendar.timegm().
>>> calendar.timegm(time.gmtime())
1293581619.0
You can turn your string into a time tuple with time.strptime(), which returns a time tuple that you can pass to calendar.timegm():
>>> import calendar
>>> import time
>>> calendar.timegm(time.strptime('Jul 9, 2009 @ 20:02:58 UTC', '%b %d, %Y @ %H:%M:%S UTC'))
1247169778
More information about calendar module here
回答 1
使用时间模块：
epoch_time = int(time.time())
Use the time module:
epoch_time = int(time.time())
回答 2
请注意，time.gmtime将时间戳映射0到1970-1-1 00:00:00。
In [61]: import time
In [63]: time.gmtime(0)
Out[63]: time.struct_time(tm_year=1970, tm_mon=1, tm_mday=1, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=3, tm_yday=1, tm_isdst=0)
time.mktime(time.gmtime(0)) 给您一个时间戳，偏移的时间取决于您的语言环境，通常可能不为0。
In [64]: time.mktime(time.gmtime(0))
Out[64]: 18000.0
与之相反的time.gmtime是calendar.timegm：
In [62]: import calendar
In [65]: calendar.timegm(time.gmtime(0))
Out[65]: 0
Note that time.gmtime maps timestamp 0 to 1970-1-1 00:00:00.
In [61]: import time
In [63]: time.gmtime(0)
Out[63]: time.struct_time(tm_year=1970, tm_mon=1, tm_mday=1, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=3, tm_yday=1, tm_isdst=0)
time.mktime(time.gmtime(0)) gives you a timestamp shifted by an amount that depends on your locale, which in general may not be 0.
In [64]: time.mktime(time.gmtime(0))
Out[64]: 18000.0
The inverse of time.gmtime is calendar.timegm:
In [62]: import calendar
In [65]: calendar.timegm(time.gmtime(0))
Out[65]: 0
回答 3
ep = datetime.datetime(1970,1,1,0,0,0)
x = (datetime.datetime.utcnow()- ep).total_seconds()
这应该与有所不同int(time.time())，但是可以安全地使用类似x % (60*60*24)
datetime-基本日期和时间类型：
与时间模块不同，日期时间模块不支持leap秒。
ep = datetime.datetime(1970,1,1,0,0,0)
x = (datetime.datetime.utcnow()- ep).total_seconds()
This should be different from int(time.time()), but it is safe to use something like x % (60*60*24)
datetime — Basic date and time types:
Unlike the time module, the datetime module does not support leap seconds.
回答 4
t = datetime.strptime('Jul 9, 2009 @ 20:02:58 UTC',"%b %d, %Y @ %H:%M:%S %Z")
t = datetime.strptime('Jul 9, 2009 @ 20:02:58 UTC',"%b %d, %Y @ %H:%M:%S %Z")
回答 5
