From: Skeets <skillet3232@...>
Date: 2006-07-31T07:15:06+09:00
Subject: Re: Howto Delete 3 Leftmost Characters

Esteban Manchado Vel�zquez wrote:
> 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.
> >
> > i grep a file and it returns the following string:
> >
> > #ip 127.0.0.1
> >
> > i now want to get rid of "#ip" so i can then strip the remaining string
> > to get rid of spaces.
> >
> > however, i can't find out how to delete the 3 leftmost characters - in
> > this case "#ip".
>
>     "#ip 127.0.0.1"[3..-1] # => " 127.0.0.1"
>     "#ip 127.0.0.1"[4..-1] # => "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+/, '') # => "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,
>
> --
> Esteban Manchado Vel�zquez <zoso@foton.es> - http://www.foton.es

Esteban, and all - thank you.

this did the trick...

if File.exist?( 'current_ip.txt' )

  f = File.open('current_ip.txt').grep(/#ip/)
  f = f[0].sub(/^#ip\s+/, '')

end

puts f

the File.open grep line returned an array.  i had to sort that out
first.  i know this file will always have only one instance of #ip - is
there any way to force grep to return as a variable instead of an
array?

also, i believe the regex Esteban gave will get rid of all white spaces
- both to the left and right of the ip address.

thanks to everyone for the help.