如何屏蔽字符串的字母和数字

Question

I have the following that works perfectly with numbers but I don't know how to get it to work with letters too.

var word = "4556364607935616";

function mask() {
  if (word.length <= 4) {
    return word;
  } else {
    var masked =
      word.substring(0, word.length - 4).replace(/\d/g, "#") +
      word.substring(word.length - 4, word.length);
    return masked;
  }
}

I'm guessing \\d targets numbers? I'm not sure where to look for a reference guide for this kind of thing. Any help would be much appreciated!

Answer 1

All you need to do is alter your regular expression slightly so that it works for letters as well. So, change /\\d/g to /[az\\d]/gi , where:

• [az\\d] matches a character in the range of a to z as well as numbers and
• the i flag ensures that both lowercase and uppercase letters are matched.

Snippet:

 var word = "4f563a46jy7u35616"; function mask() { if (word.length <=4) { return word } else { var masked = word.substring(0, word.length - 4).replace(/[az\\d]/gi,"#") + word.substring(word.length - 4, word.length) return masked; } } console.log(mask(word));

Answer 2

The effective solution would be to not mask the string, but build a new string with provided length

var word = "4556364607935616";

function mask() {
  var LENGTH = 4
  var CHAR = '#'
  if (word.length <= LENGTH) {
    return word
  }
  var leftSideLength = word.length - LENGTH
  var result = ''
  while (leftSideLength--) result += CHAR
  return result + word.substring(word.length - 4, word.length)
}

Below the performance advantage shown

 var word = "4556364607935616"; function maskNotRegex() { var LENGTH = 4 var CHAR = '#' if (word.length <= LENGTH) { return word } var leftSideLength = word.length - LENGTH var result = '' while (leftSideLength--) result += CHAR return result + word.substring(word.length - 4, word.length) } // Credit: Angel Politis's anwer on this Post function maskRegex() { if (word.length <= 4) { return word } else { var masked = word.substring(0, word.length - 4).replace(/[az\\d]/gi, "#") + word.substring(word.length - 4, word.length) return masked; } } // Performance test // Credit: https://stackoverflow.com/a/17943511/2308005 var iterations = 1000000; console.time('Using regex'); for (var i = 0; i < iterations; i++) { maskRegex(word); }; console.timeEnd('Using regex') console.time('Not using regex'); for (var i = 0; i < iterations; i++) { maskNotRegex(word); }; console.timeEnd('Not using regex')

To get more info about regex performance, read codinghorror post

Answer 3

RegExp Reference Guide with Tutorials : http://www.regular-expressions.info/

RegExp Playground : https://regexr.com/ (online tool to learn, build, & test regular expressions)

If you need to avoid replacing dashes - etc. use the word character \\w (includes underscore):

 var word="abc-364-079-5616", masked=word.replace(/\\w/g, "#"); //[a-zA-Z0-9_] if(word.length > 4) { masked = masked.substring(0, word.length-4) + word.substring(word.length-4); } else { masked = word; } console.log(masked); // "###-###-###-5616"

Here's how masked changes in the last example:

• masked = "###-###-###-####"
• masked = "###-###-###-5616"

No RegExp Example

Here's an example that doesn't use regular expressions (masks any character):

 var word = "abc6364607935616", masked = word; // word & masked are the same if(word.length > 4) { masked = new Array(word.length - 4).join('#'); // create 4 less than word masked += word.substring(word.length - 4); // add on 4 last characters } console.log(masked); // "###########5616"

Here's how masked changes in the last example:

• masked = "abc6364607935616"
• masked = "############"
• masked = "############5616"

Answer 4

This is the shortest I got. You can use Regex pattern ( \\w(?=\\w{4}) ) to mask the last 4 letters like so:

 let word = "4f563a46jy7u35616"; function mask() { if (word.length <= 4) { return word; } else { return word.replace(/\\w(?=\\w{4})/g, "#"); } } console.log(mask(word));