如何在gmtime()的时间+日期输出之后获得秒数?

How do you do reverse gmtime(), where you put the time + date and get the number of seconds?

如何进行反向gmtime()，在这里输入时间+日期并获得秒数?

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.

我有像“2009年7月9日@ 20:02:58 UTC”这样的字符串，我想要返回从2009年7月9日开始的秒数。

I have tried time.strftime but I don't know how to use it properly, or if it is the correct command to use.

我尝试过时间。但是我不知道如何正确地使用它，或者它是否是正确的命令。

6 个解决方案

#1

100

You want calendar.timegm().

你想要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():

可以将字符串转换为带有time.strptime()的时间元组，该时间元组返回一个可以传递给calendar.timegm()的时间元组:

>>> 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

更多有关日历模块的信息请参见这里

#2

450

Use the time module:

利用这段时间模块:

epoch_time = int(time.time())

#3

Note that time.gmtime maps timestamp 0 to 1970-1-1 00:00:00.

注意时间。gmtime地图时间戳0到1970-1-1 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.

mktime(time.gmtime(0)))为您提供一个时间戳，该时间戳的偏移量取决于您的语言环境，通常可能不是0。

In [64]: time.mktime(time.gmtime(0))
Out[64]: 18000.0

The inverse of time.gmtime is calendar.timegm:

时间的倒数。gmtime calendar.timegm:

In [62]: import calendar    
In [65]: calendar.timegm(time.gmtime(0))
Out[65]: 0

#4

t = datetime.strptime('Jul 9, 2009 @ 20:02:58 UTC',"%b %d, %Y @ %H:%M:%S %Z")

#5

There are two ways, depending on your original timestamp:

根据您的原始时间戳，有两种方法:

mktime() and timegm()

mktime()和timegm()

http://docs.python.org/library/time.html

#6

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)

这应该与int(time.time())不同，但是使用x %(60*60*24)之类的东西是安全的

datetime — Basic date and time types:

日期-基本日期和时间类型:

Unlike the time module, the datetime module does not support leap seconds.

与time模块不同，datetime模块不支持闰秒。

#1

100