Convert apache log date/time to sql

hallikpapa has asked for the wisdom of the Perl Monks concerning the following question:

I have an apache log that stores date/time like this (which is also how it looks in my $var; square brackets and all):

[26/Mar/2009:12:12:48 -0500]
[download]

I am attempting to use Date::Parse.
http://search.cpan.org/~gbarr/TimeDate-1.16/lib/Date/Parse.pm

But it doesn't seem to support the format I am in. I am horrible and regular expressions and need some help/suggestions at converting the date above.

This is so I can compare dates from a table that stores dates like:
2009-03-27 14:18:40
<edit> Nevermind. I made a mistake. Didn't mean to post this, sorry. Can delete it.

Comment on Convert apache log date/time to sql Download Code

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Re: Convert apache log date/time to sql by codeacrobat (Chaplain) on Mar 31, 2009 at 20:28 UTC
For the curious, this is one way to parse the date. `TZ=UTC perl -MDate::Manip=UnixDate -e ' $_=q{[2009/Mar/26:12:12:48 -0500]}; s/:/ /; tr/[]//d; print UnixDate($_,"%Y-%m-%d %H:%M:%S");' 2009-03-26 17:12:48` [download] `print+qq(\L@{[ref\&@]}@{['@'x7^'!#2/"!4']});`	[reply] [d/l] [select]