From: "Esteban Manchado Velázquez" <zoso@...>
Date: 2006-07-31T06:51:35+09:00
Subject: Re: Howto Delete 3 Leftmost Characters

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On Mon, Jul 31, 2006 at 06:30:13AM +0900, Skeets wrote:
> i'm sure this is easy, but i've gone through Pickaxe's string methods,
> searched the web and searched the groups, and i can't figure out how to
> do this in Ruby.
>=20
> i grep a file and it returns the following string:
>=20
> #ip 127.0.0.1
>=20
> i now want to get rid of "#ip" so i can then strip the remaining string
> to get rid of spaces.
>=20
> however, i can't find out how to delete the 3 leftmost characters - in
> this case "#ip".

    "#ip 127.0.0.1"[3..-1] # =3D> " 127.0.0.1"
    "#ip 127.0.0.1"[4..-1] # =3D> "127.0.0.1"

But you're probably better off using regular expressions instead of fixed
indices:

    "#ip 127.0.0.1".sub(/^#ip\s+/, '') # =3D> "127.0.0.1"

That is, "remove, from the beginning of the line, '#ip' followed by one or
more space characters (be them spaces, tabs or whatever)". If you don't know
regular expressions, _and_ you're into text processing, I recommend you to =
go
and read some book about regular expressions and practice a little (under
Linux there are a couple of handy utilities for that; I'm sure there must be
also for other platforms).

    Regards,

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

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