From: w_a_x_man <w_a_x_man@...>
Date: 2010-10-26T23:50:15+09:00
Subject: Re: Change captured values in a regexp

On Oct 26, 8:47 am, Bruno Sousa <brgso...@gmail.com> wrote:
> Hi everyone!
> Supposing there is a string like this:
> Oct 22 06:25:35 machine kernel: [271182.956004] IN= OUT=eth0
> SRC=10.8.0.248 DST=192.5.5.241
>
> I am using the following regexp to capture time:
> /Oct (\d{2}) (\d{2}:\d{2}:\d{2})/
>
> How can I alter the hour value only and keep it in the result returned?
> I want to decrease 3 hours.
> Instead of #<MatchData "Oct 22 06:25:35" 1:"22" 2:"06:25:35">
> It should return  #<MatchData "Oct 22 06:25:35" 1:"22" 2:"03:25:35">
>
> --
> Posted viahttp://www.ruby-forum.com/.

What is going to happen when you subtract 3 hours from a date
like this:
Oct 01 02:25:35

require 'date'
    ==>true
t=DateTime.parse("Oct 01 02:25:35") - Date.time_to_day_fraction(3,0,0)
    ==>#<DateTime: 42430529827/17280,0,2299161>
t.strftime("%b %d %H:%M:%S")
    ==>"Sep 30 23:25:35"