Python 时间老化,第 2 部分:时区
问题描述
继我之前的问题之后,Python 老化时间,我现在遇到了一个关于时区的问题,事实证明它并不总是+0200".所以当 strptime 试图解析它时,它会抛出一个异常.
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.
我曾想过用 [:-6] 或其他什么来切断 +0200,但有没有真正的方法可以用 strptime 做到这一点?
I thought about just chopping off the +0200 with [:-6] or whatever, but is there a real way to do this with strptime?
如果重要的话,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
解决方案
2.6 版中的新功能.
New in version 2.6.
对于一个简单的对象,%z 和 %Z格式代码替换为空字符串.
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,我认为您必须手动解析它.
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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