From: Glen Holcomb <damnbigman@...>
Date: 2011-08-11T20:52:26+09:00
Subject: Re: Odd regexp behavior

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On Thu, Aug 11, 2011 at 12:41 AM, 7stud -- <bbxx789_05ss@yahoo.com> wrote:

> Also,
>
>
> str =3D "hello\nhello"
>
> str.scan(/^hell/) {|match| p match}
> puts "-" * 20
> str.scan(/\Ahell/) {|match| p match}
>
> --output:--
> "hell"
> "hell"
> --------------------
> "hell"
>
> --
> Posted via http://www.ruby-forum.com/.
>
>
Hmmm... maybe I should have posted from the beginning rather than where I
had gotten to in my attempt to solve my problem.

I am tyring to parse log files and am pulling out encapsulated fields so
that I can split the line in a sane way.  I have the following regex which =
I
am using to do that:

/\s(".*?")(\s|$)/

Now as to why I'm looking for the \s before and the \s or $ after.  It turn=
s
out that some of the user agent strings are in a format like "\"Custom
Agent\"=3D\"Mozilla ...\""\n

I had been using the regex Ryan suggested earlier until I discovered the
nested quotes.

The expression above works for quoted strings surrounded by spaces but not
the last one on the line.  I've tried changing $ to \n and that didn't make
any difference.

Here is the exact code I'm using:

x =3D "encapsulatorhere"
stash =3D {}

gen_encap_matches.each do |encapex|
line.gsub!(encapex) do |match|
x.next!
stash[x] =3D $1
@job.log_format.separator + x + @job.log_format.separator
end
end

gen_encap_matches just creates the regexes from a list of encapsulation
characters.

I'm at a complete loss as to why it won't grab the last quoted string in th=
e
line.

--=20
"Hey brother Christian with your high and mighty errand, Your actions speak
so loud, I can=92t hear a word you=92re saying."

-Greg Graffin (Bad Religion)

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