From: Paolo Negri <hungrylist@...>
Date: 2007-07-11T08:05:33+09:00
Subject: Re: Help: Efficient regular expression

sorry for being OT since I'm not going to talk about ruby or regexp

If the string you're parsing is an output from the ps command you can
simplify your life using the -o option that prints only the fields you
need.

I.E. in gnu Linux

ps -ao pid,command

just outputs pid and command columns.
Be careful since the command column can contain spaces.


Paolo

On 10/07/07, Divya Badrinath <dbadrinath@dash.net> wrote:
> string = "root     14051 14033  3 08:39 pts/2    00:00:00 /bin/bash"
>
> i need to fetch 14051 and /bin/bash from the string
>
> can someone  help me to write an efficient regular expression for that.
>
> i am a beginner, i wrote
> string =~ /(\d+)\s+(\d+)\s+\d+\s+\d+:\d+\s+.*\s+\d+:\d+:\d+\s+(.*)\s/
>
> i know this is not the efficient way of doing it.
>
> Please help.
>
> --
> Posted via http://www.ruby-forum.com/.
>
>