3

I have a script that is parsing out fields within email headers that represent dates and times. Some examples of these strings are as follows:

Fri, 10 Jun 2011 11:04:17 +0200 (CEST) Tue, 1 Jun 2011 11:04:17 +0200 Wed, 8 Jul 1992 4:23:11 -0200 Wed, 8 Jul 1992 4:23:11 -0200 EST

Before I was confronted with the CEST/EST portions at the ends of some the strings I had things working pretty well just using datetime.datetime.strptime like this:

msg['date'] = 'Wed, 8 Jul 1992 4:23:11 -0200' mail_date = datetime.datetime.strptime(msg['date'][:-6], '%a, %d %b %Y %H:%M:%S')

I tried to put a regex together to match the date portions of the string while excluding the timezone information at the end, but I was having issues with the regex (I couldn't match a colon).

Is using a regex the best way to parse all of the examples above? If so, could someone share a regex that would match these examples? In the end I am looking to have a datetime object.

Improve this question

2 Answers 2

Reset to default
7

From python time to age part 2, timezones:

from email import utils utils.parsedate_tz('Fri, 10 Jun 2011 11:04:17 +0200 (CEST)') utils.parsedate_tz('Fri, 10 Jun 2011 11:04:17 +0200') utils.parsedate_tz('Fri, 10 Jun 2011 11:04:17')

The output is:

(2011, 6, 10, 11, 4, 17, 0, 1, -1, 7200) (2011, 6, 10, 11, 4, 17, 0, 1, -1, 7200) (2011, 6, 10, 11, 4, 17, 0, 1, -1, None)
Improve this answer
Sign up to request clarification or add additional context in comments.

1 Comment

I saw that the old rfc822 module had similar functionality but I was not aware of email.utils. Thank you.
2

Perhaps I misunderstood your question, but wont a simple split suffice?

#!/usr/bin/python d = ["Fri, 10 Jun 2011 11:04:17 +0200 (CEST)", "Tue, 1 Jun 2011 11:04:17 +0200", "Wed, 8 Jul 1992 4:23:11 -0200", "Wed, 8 Jul 1992 4:23:11 -0200 EST"] for i in d: print " ".join(i.split()[0:5]) Fri, 10 Jun 2011 11:04:17 Tue, 1 Jun 2011 11:04:17 Wed, 8 Jul 1992 4:23:11 Wed, 8 Jul 1992 4:23:11
Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.