Python老化时间，第2部分：时区 [英] Python time to age, part 2: timezones

问题描述

接着我的上一个问题， Python的生存时间 ，我现在遇到了有关时区的问题，事实证明，时区并不总是为 +0200。因此，当strptime试图这样解析它时，它会引发异常。

Following on from my previous question, Python time to age, I have now come across a problem regarding the timezone, and it turns out that it's not always going to be "+0200". So when strptime tries to parse it as such, it throws up an exception.

我考虑过用[：-6]或其他任何东西来截断+0200，但是

I thought about just chopping off the +0200 with [:-6] or whatever, but is there a real way to do this with strptime?

如果重要，我正在使用Python 2.5.2。

I am using Python 2.5.2 if it matters.

>>> from datetime import datetime
>>> fmt = "%a, %d %b %Y %H:%M:%S +0200"
>>> datetime.strptime("Tue, 22 Jul 2008 08:17:41 +0200", fmt)
datetime.datetime(2008, 7, 22, 8, 17, 41)
>>> datetime.strptime("Tue, 22 Jul 2008 08:17:41 +0300", fmt)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.5/_strptime.py", line 330, in strptime
    (data_string, format))
ValueError: time data did not match format:  data=Tue, 22 Jul 2008 08:17:41 +0300  fmt=%a, %d %b %Y %H:%M:%S +0200

推荐答案

2.6版中的新功能。

New in version 2.6.

一个天真的对象，％z和％Z
格式代码将替换为空的
字符串。

For a naive object, the %z and %Z format codes are replaced by empty strings.

看起来这仅在> = 2.6中实现，并且我认为您必须手动对其进行解析。

It looks like this is implemented only in >= 2.6, and I think you have to manually parse it.

除了删除时区数据，我看不到其他解决方案：

I can't see another solution than to remove the time zone data:

from datetime import timedelta,datetime
try:
    offset = int("Tue, 22 Jul 2008 08:17:41 +0300"[-5:])
except:
    print "Error"

delta = timedelta(hours = offset / 100)

fmt = "%a, %d %b %Y %H:%M:%S"
time = datetime.strptime("Tue, 22 Jul 2008 08:17:41 +0200"[:-6], fmt)
time -= delta

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