I'd use a global search and replace command, if it were me, and I was using
sed: sed -ie 's/&pg=[^&]//g' lindsay.html

On 13 July 2010 18:13, Lindsay Holmwood <[email protected]> wrote:

> Now you've got the search, I'm curious how you are going to do the replace.
>
> Is the Perlism to just use the substitute operator, or split on the
> pattern, iterate through the array, and join again?
>
> Lindsay
>
> On 14 July 2010 10:30, Jamie Wilkinson <[email protected]> wrote:
> > Try:
> >
> > /&pg=[^&]*/
> >
> > match zero or more of the character class that is not an ampersand.
> >
> > On 13 July 2010 17:21, Peter Rundle <[email protected]> wrote:
> >
> >> Hi Sluggers,
> >>
> >> I'm sure some of you genii have a real quick solution to this.
> >>
> >> I'm trying to find and replace and argument in a url. The url is of the
> >> form
> >>
> >> &pg=something&arg=somethingelse
> >>
> >>
> >> I want to take out the &pg=something but the "&arg=" may or may not be
> >> there. How do I say match the &pg=something up to but not including the
> next
> >> & (which may or may not be there).
> >>
> >>        "/&pg=.*&/"
> >>
> >> But also I think & is a special char (no?) that means "put the matched
> bit
> >> back", though is that only on the replace side? (my question relates
> >> strictly to the matching side).
> >>
> >>
> >> TIA's
> >>
> >> Pete
> >>
> >>
> >> --
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>
>
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