Re: Capturing part of regex search

On Sun, Aug 23, 2015 at 6:52 PM, Aziz Saleh <azizsaleh@gmail.com> wrote: > > > On Sun, Aug 23, 2015 at 6:30 PM, Ian Evans <dheianevans@gmail.com> wrote: > >> I've been looking at the regex functions. Either it's a lack of caffeine >> or >> it's right in front of my nose, but how do I put part of a regex find into >> a variable? Plus I suck at regex. >> >> e.g. >> >> I'm interested in capturing the names of people who are found in text >> surrounded by the following format: >> >> "John Doe":/cr/johndoe/ or "Jane Smith":/cr/janesmith/ >> >> I'd want to place John Doe or Jane Smith into the variable. The search >> will >> always be for "name":/cr/nameurl/ >> >> What regex and php function would I use? Thanks for any pointers. One of >> these days I'll master regex. >> > > If the names will always contain double quotes, it should be as easy as: > > $var = ' "John Doe":/cr/johndoe/ or "Jane Smith":/cr/janesmith/'; > preg_match_all('/".*"/U', $var, $res); > print_r($res); > > U to make it ungreedy (its greedy by default). If there are other quotes > in the text, you can add the colon to the expression as well. > Thanks. There might be other quotes in the article, so if I include the colon, it'd be before the /U?