用 # 替换所有字符，最后 4 个字符除外

Question

function maskify(cc) {
    var dd = cc.toString();
    var hash = dd.replace((/./g), '#');
    for (var i = (hash.length - 4); i < hash.length; i++) {
        hash[i] = dd[i];
    }
    return hash;
}

I am trying to replace all chars with # except for last 4. Why isn't it working?

Answer 1

You could do it like this:

dd.replace(/.(?=.{4,}$)/g, '#');

 var dd = 'Hello dude'; var replaced = dd.replace(/.(?=.{4,}$)/g, '#'); document.write(replaced);

Answer 2

If you find the solution, try this trick

function maskify(cc) {
  return cc.slice(0, -4).replace(/./g, '#') + cc.slice(-4);
}

Answer 3

To replace a character in a string at a given index, hash[i] = dd[i] doesn't work. Strings are immutable in Javascript. See How do I replace a character at a particular index in JavaScript? for some advice on that.

Answer 4

尝试这个：

return cc.replace(/.(?=.{4})/g, "#");

Answer 5

function maskify(cc) {
  let arr = cc.split('');
  for (let i = 0; i < arr.length - 4; i++){
    arr[i] = '#';
  }
  return arr.join('');
}