From: Todd Benson <caduceass@...>
Date: 2007-07-11T08:50:46+09:00
Subject: Re: Help: Efficient regular expression

On 7/10/07, Divya Badrinath <dbadrinath@dash.net> wrote:
> Robert Dober wrote:
> > On 7/10/07, James Edward Gray II <james@grayproductions.net> wrote:
> >>    sec, last = cols.values_at(1, -1)
> > Very interesting James, I seem to be rather extreme and
> >
> >   sec, last =  string.split.values_at(1, -1)
> > might be a tad to long for one line in your style, however Ruby syntax
> > just supports this marvelous syntax :)
> >
> >    sec, last = string.split.
> >                        values_at(1, -1)
> >
> > Robert
>
> cmd = string[/\s(\S+)$/, 1]
> doesnt fetch me anything:)
>
> program=string.split.last
> what if
> string = "root     14051 14033  3 08:39 pts/2    00:00:00 /bin/bash -x
> -s"
> it fetches only -s for me.
> sec, last =  string.split.values_at(1, -1)
> doesnt work for the same reason
> i need everything after 00.00.00 till the end
> i.e., /bin/bash -x -s
>
> program=string[/[a-z\/]+$/]
> the command column mauy start with character. i dont want to limit it in
> my regexp. it has to be generic.
>
> with all your comments, i tried
>       pid = run_process[/\s(\d+)/, 1]
>       cmd = run_process[/:\d+:\d+\s(\S.*)\s$/, 1]
>
> is there any other way?

It's not fancy, but I'll throw it in:

s = 'root     14051 14033   3 08:39 pts/2    00:00:00 /bin/bash -x -s'
_, pid, _, cmd = *(s.match /(\d+)\s.*(:\d+){2}\s(.*?)$/)

so, if you're using a hash like I think you might be:

s = <output of ps command>
h = {}
s.each_line do |line|
  _, pid, _, cmd = *(line.match /(\d+)\s.*(:\d+){2}\s(.*?)$/)
  h[pid] = cmd
end

I think that should work.

Todd

_, pid