I've taken to using `match` for this. Your example: (regexp-match #rx"\"(.*)" "a\"b")
I would write as: (match "a\"b" [(regexp "\"(.*)" (list _ x)) x]) ; "b" Handling the no-match case by returning #f: (match "ab" [(regexp "\"(.*)" (list _ x)) x] [_ #f]) ; #f Laurent's approach of defining a function works well, too. But I often find myself matching multiple items, and use the list to bind to multiple variables: (match "x=y" [(pregexp "(.+)\\s*=\\s*(.+)" (list _ k v)) (values k v)] [_ #f]) ; "x" ; "y" So I like match as it works in the general case. Note that there's `pregexp` as well as `regexp`, like I used here. On Fri, May 10, 2013 at 11:27 AM, Don Green <[email protected]> wrote: > Regexp question: > Is there a way using regexp only to return a list with a single element? > I could use a Racket list function such as caar to return the second element > but I am wondering if there is a way to do this using regexp only. > > For example this regexp-match function generates the "b" that I want but it > is the second element in the list. It would be ideal, from my perspective, > if that was all it generated. Is there a way to write the regexp-match > expression so that '("b") is output? > > I get this: > (regexp-match #rx"\"(.*)" "a\"b") ; => '("\"b" "b") > > I'd prefer: > (regexp... "a\"b" ) ; => '("b") > > Thanks. > Don Green > > ____________________ > Racket Users list: > http://lists.racket-lang.org/users > ____________________ Racket Users list: http://lists.racket-lang.org/users
