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 >> >> >> -- >> SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ >> Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html >> > -- > SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ > Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html > -- w: http://holmwood.id.au/~lindsay/ t: @auxesis -- SLUG - Sydney Linux User's Group Mailing List - http://slug.org.au/ Subscription info and FAQs: http://slug.org.au/faq/mailinglists.html
