View Problem
Replace all regex matches in a string with a dynamic string
Transform 
scala
erlang
Submit a new solution for scala or erlang
There are 13 other solutions in additional languages (clojure, cpp, fantom, fsharp ...)
"The {Quick} Brown {Fox}" into "The kciuQ Brown xoF" by reversing words in braces using the regex /\{(\w+)\}/.
val m = Pattern.compile("\\{(\\w+)\\}").matcher("The {Quick} Brown {Fox}")
val sb = new StringBuffer(32) ; val rsb = new StringBuffer(8)
while (m.find) { rsb.replace(0, rsb.length, m.group(1)) ; m.appendReplacement(sb, rsb.reverse.toString) }
m.appendTail(sb)
val sb = new StringBuffer(32) ; val rsb = new StringBuffer(8)
while (m.find) { rsb.replace(0, rsb.length, m.group(1)) ; m.appendReplacement(sb, rsb.reverse.toString) }
m.appendTail(sb)
% Erlang regular expressions lack both group capture and backreferences, thus this problem is not directly
% solvable. Presented solution is close, but not on-spec
String = "The {Quick} Brown {Fox}",
{match, FieldList} = regexp:matches(String, "\{([A-Za-z0-9_]+)\}"),
NewString = lists:foldl(fun ({Start, Length}, S) -> replstr(S, lists:reverse(string:substr(S, Start, Length)), Start) end, String, FieldList),
% solvable. Presented solution is close, but not on-spec
String = "The {Quick} Brown {Fox}",
{match, FieldList} = regexp:matches(String, "\{([A-Za-z0-9_]+)\}"),
NewString = lists:foldl(fun ({Start, Length}, S) -> replstr(S, lists:reverse(string:substr(S, Start, Length)), Start) end, String, FieldList),
Submit a new solution for scala or erlang
There are 13 other solutions in additional languages (clojure, cpp, fantom, fsharp ...)
