regex - JavaScript - RegExp - Replace useless parentheses in string -

regex - JavaScript - RegExp - Replace useless parentheses in string -

so have string this, ready evaluated:

"getinfo(((((2+2)*(3/4) / ((44))))))"

so, there's 1 place tripple parenthese is useless , 1 place double 1 useless, string can simplyfied this:

"getinfo((2+2)*(3/4) / (44))"

in order sort these unnesseccary parentheses out can replace every 2 useless parentheses 1 parenthese, this:

"do((((2+2 - 3)*(2))))" -> "do(((2+2 - 3)*(2)))" -> "do((2+2 - 3)*(2))"

is there anyway either regexp or looping using string.replace method this, , how?

this seems work:

function simplify(str) { var i=0; homecoming (function recur(s, b) { var c = str.charat(i++); // next char if(!c || c == ')') homecoming s; // end of string or end of inner part if(c == '(') { var s1 = recur('', true), // inner part s2 = recur(''); // next part homecoming s + (!b || s2 ? '('+s1+')' : s1) + s2; } homecoming recur(s+c); // go on next char })(''); }

this code should equivalent, less function calls:

function simplify(str) { var i=0; homecoming (function recur(b) { var c, s = ''; while(c = str.charat(i++)) { // maintain getting chars if(c == ')') homecoming s; // end of inner part if(c == '(') { var s1 = recur(true), // inner part s2 = recur(); // next part homecoming s + (!b || s2 ? '('+s1+')' : s1) + s2; } s += c; // add together current char b = false; } homecoming s; })(); }

the part gets rid of unnecessary parentheses one:

return s + (!b || s2 ? '('+s1+')' : s1) + s2;

basically, returns current string, concatenated inner part (maybe parenthesized), concatenated next part.

the inner part parenthesized either if next part not empty, or if b falsy (that is, if inner part not wrapped in parentheses).

note codes above expect well-formed expression. if not,

the returned string truncated if there ) without corresponding (

simplify("a)b"); // "a"

the returned string have additional ) @ end if there unclosed (:

simplify("a(b"); // "a(b)"

javascript regex string