From: Amishera Amishera <amishera2007@...>
Date: 2010-04-01T07:11:33+09:00
Subject: Re: how to solve this regular expression problem

Amishera Amishera wrote:
> I have strings of this form:
> 
> s = 'C#039;era una volta il east'
> 
> I want to replace '#039;' with the octal representation of 39 ie 47 so
> the string becomes
> 
> s = 'C\47era una volta il east'
> 
> I have tried this:
> 
>   1 def conv_char(num)
>   2         puts num.to_i.to_s(8)
>   3         num.to_i.to_s(8)
>   4 end
>   5
>   6 if __FILE__ == $0
>   7         s = 'C#039;era'
>   8         s.gsub!(/#(\d+);/, conv_char("\1"))
>   9         puts s
>  10 end
> 
> The outptu i get is
> 0
> 'C#0era una volta il east'
> 
> I even tried
>  8         s.gsub!(/#(\d+);/, conv_char('\1'))
>  8         s.gsub!(/#(\d+);/, conv_char($1))
> 
> nothing seems to work.
> 
> how to solve this gracefully? I guess I can do that using double
> replacement like collecting the matched patterns (#039;) using scan
> method and then converting and doing another gsub using a for loop. But
> that seems to much for the poor regular expression engine.

I can do this also:

s.gsub!(/#(\d+);/) { puts $1; conv_char($1) }

But I was wondering how to use both the \1 syntax and the function call 
together in the gsub call in the first approach.
-- 
Posted via http://www.ruby-forum.com/.