Call me crazy! s/(&|?)pg=[^&]*/\1/
(correct escaping of & and ? left as an exercise for someone actually using this :) On 27 July 2010 08:03, Martin Barry <[email protected]> wrote: > Sorry to bring up an old thread but I just had to comment on this... > > $quoted_author = "Jamie Wilkinson" ; > > > > Try: > > > > /&pg=[^&]*/ > > > > match zero or more of the character class that is not an ampersand. > > Except there is nothing stopping the variables being reordered, no? So you > may need to match a leading ? instead of &. > > You could get crazy and try to do this in a single regex but two stage is > clearer. e.g. > > sed -e 's/&pg=[^&]*//g' -e 's/?pg=[^&]*&/?/' > > > cheers > Marty > -- > 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
