From: "Esteban Manchado Velázquez" <zoso@...>
Date: 2006-08-01T00:38:25+09:00
Subject: Re: Howto Delete 3 Leftmost Characters

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On Mon, Jul 31, 2006 at 07:25:04AM +0900, Skeets wrote:
> > also, i believe the regex Esteban gave will get rid of all white spaces
> > - both to the left and right of the ip address.
>=20
> actually, this isn't true so i  still have to use the strip method.
>=20
> i have some regex experience - but not too much.
>=20
> i pretty much take some pain killers and then try and figure out what i
> need as the need arises.  ;-)
>=20
> i'll bet this regex can be update to remove all spaces, however, i will
> have to look into that next week.

    Something like:

    s =3D "#ip   127.0.0.1    "
    s.sub(/^#ip\s+(.+?)\s*$/, '\1')

That's way uglier, because you have to match the whole line and then
substitute everything for the first parens contents.... _and_ you have to u=
se
".+?" instead of simply ".+" inside them, or otherwise you will get everyth=
ing
until the end of the line :-/

    The "?" after "*" or "+" means non-greediness, BTW.

--=20
Esteban Manchado Vel=E1zquez <zoso@foton.es> - http://www.foton.es
EuropeSwPatentFree - http://EuropeSwPatentFree.hispalinux.es

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