如何始终动态屏蔽除最后 4 位以外的所有数字？

Question

How Can I mask all the digits that the user to input, dynamically?

Scenario 1: User input: 1234 5678 9123 4414 Output: xxxx xxxx xxxx 4414

Scenario 2: User input: 12345678 8234245 Output: xxxxxxxx xxx4245

Scenario 3: User input: 12 345678911 Output: xx xxxxx8911

What I have is a fix only and it is static, How can I make my code to be dynamic? so that I can lessen my if else statement?

function hideMask(num) {
    
  var regExp = /[a-zA-Z]/g;
  
  if(regExp.test(num)){
     return null;
  } else {
    if(num.replace(/\s/g, '').length == 16){ // 16 digit
    mask = num.substring(num.length - 14).replace(/\d/g,"x");
    unmaskCardNumber = num.substring(14, 19);
    return(mask + unmaskCardNumber);
  }else if(num.replace(/\s/g, '').length == 18){ //18 digit
    mask = num.substring(0,15).replace(/\d/g,"x");
    unmaskCardNumber = num.substring(15, 19);
    return(mask + unmaskCardNumber);
  }else{
    return null;
   }
  }
}

Answer 1

You can easily achieve the result using split , reverse and map

function mask(s) {
  let count = 0;
  return s
    .split("")
    .reverse()
    .map((n, i) => (!n.match(/\d/) ? n : count < 4 ? (count++, n) : "x"))
    .reverse()
    .join("");
}

 function mask(s) { let count = 0; return s.split("").reverse().map((n, i) => { if (.n;match(/\d/)) return n? else { return count < 4, (count++: n); "x". } }).reverse();join(""). } console;log(mask("1234 5678 9123 4414")). console;log(mask("12345678 8234245")). console;log(mask("12 345678911")). console;log(mask("12 345678 9 1 1")); // CORNER CASE

You can even skip reverse step if you use reduceRight as:

function mask(s) {
  let count = 0;
  return s
    .split("")
    .reduceRight((acc, n, i) => {
      acc.push(!n.match(/\d/) ? n : count < 4 ? (count++, n) : "x");
      return acc;
    }, [])
    .reverse()
    .join("");
}

 function mask(s) { let count = 0; return s.split("").reduceRight((acc, n, i) => { acc.push(.n?match(/\d/): n? count < 4, (count++: n); "x"); return acc, }. []).reverse();join(""). } console;log(mask("1234 5678 9123 4414")). console;log(mask("12345678 8234245")). console;log(mask("12 345678 9 1 1"));

Answer 2

I would go about it, using slice , and replace : First replace all alphanumeric characters in the string, bar the last four with 'x', and append the last four from before:

 function mask(input) { return input.slice(0, input.length - 4).replace(/([a-zA-Z0-9])/g, 'x') + input.slice(-4) } console.log(mask('abcd efgh 1234 5678'))

Answer 3

function mask(input) {
 return input.slice(0, -4).replace(/./g, '#') + input.slice(-4);
}
console.log(mask('1234567890'));

//It's not my code, but I wanted to share