find first matches character in regular expression

Question

i have a string that i want to find "a" character and first "c" character and replace with "*" character, but regular expression find all "a" and "c" character i don't know how do it.

here my code:

var pattern=/[a][c]/g; //here my pattern
var str1="aaaaaaaaaabbbbbbaaaaaaaaaabbbccccccccccbbbbbbbbccc";
var rep=str1.replace(pattern,"*");
$("p").html(rep);

remove global `/g` flag from regex, use `str1.replace('a', '*').replace('c', '*')` — AZ_, Nov 09 '19 at 18:31
@AZ_ i want replace both characters "c" and "a" when i removed global just replace "a" character — , Nov 09 '19 at 18:47

Nina Scholz · Answer 1 · 2019-11-09T18:38:37.710

You need two replacements, because a single one replaces either a or c, depending on the first occurence in the string.

var string = "aaaaaaaaaabbbbbbaaaaaaaaaabbbccccccccccbbbbbbbbccc",
    result = string
        .replace(/a/, '*')
        .replace(/c/, '*');

console.log(result);

An approach with a single replace and a closure over a hash table.

var string = "aaaaaaaaaabbbbbbaaaaaaaaaabbbccccccccccbbbbbbbbccc",
    result = string.replace(/[ac]/g, (h => c => h[c] ? c : (h[c] = '*'))({}));

console.log(result);

score 0 · Answer 2 · answered Nov 09 '19 at 18:35

You can use capture groups for all a or c sequences but the 1st, and replace them with '*$1' ($1 is the replacement pattern for the capture group):

const str = `aaaaaaaaaabbbbbbaaaaaaaaaabbbccccccccccbbbbbbbbccc`;

const result = str.replace(/[ac]([ac]+)/g, '*$1');

console.log(result);

find first matches character in regular expression

2 Answers2