javascript - 我需要使用字母数字校验和来验证输入字符串

Question

我有以下函数，可以根据 Luhn 算法验证仅包含数字的数字输入:

function isCheckdigitCorrect(value) {
// accept only digits, dashes or spaces
  if (/[^0-9-\s]+/.test(value)) return false;

  var nCheck = 0, nDigit = 0, bEven = false;
  value = value.replace(/\D/g, "");

  for (var n = value.length - 1; n >= 0; n--) {
    var cDigit = value.charAt(n),
      nDigit = parseInt(cDigit, 10);

    if (bEven) {
      if ((nDigit *= 2) > 9) nDigit -= 9;
    }

    nCheck += nDigit;
    bEven = !bEven;
  }

  return (nCheck % 10) == 0;
}

无论如何，我也可以验证字母数字，所以假设我有一个有效的 ID: AC813(6) ， () 是校验和。那么有没有一种方法可以防止用户错误地输入 AF813(6)，这样就会告诉用户错误的 ID。

感谢您的帮助

Answer 1

用数字替换字母字符来计算校验和会严重降低检查的稳健性，我能想到的最简单的建议是使用维基百科上描述的Luhn mod N algorithm。

将算法转换为 JavaScipt 相对简单:以下不是我的代码，而是 wiki 文章的翻译 - 所以我不会假装它是最佳的。它旨在处理不区分大小写的 ASCII 字母字符和十进制数字的字符串。有关文档，请参阅 wiki。

// based on https://en.wikipedia.org/wiki/Luhn_mod_N_algorithm

var charset = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
function NumberOfValidInputCharacters () { return charset.length; }
function CodePointFromCharacter(character) { return charset.indexOf(character)};
function CharacterFromCodePoint( codePoint) { return charset[codePoint]};

function GenerateCheckCharacter (input) {

    var factor = 2;
    var sum = 0;
    var n = NumberOfValidInputCharacters();
    input = input.toUpperCase();

    // Starting from the right and working leftwards is easier since 
    // the initial "factor" will always be "2" 
    for (var i = input.length - 1; i >= 0; i--) {
        var codePoint = CodePointFromCharacter(input[i]);
        if( codePoint < 0) {
            return "";
        }

        var addend = factor * codePoint;

        // Alternate the "factor" that each "codePoint" is multiplied by
        factor = (factor == 2) ? 1 : 2;

        // Sum the digits of the "addend" as expressed in base "n"
        addend = Math.floor(addend / n) + (addend % n);
        sum += addend;
    }

    // Calculate the number that must be added to the "sum" 
    // to make it divisible by "n"
    var remainder = sum % n;
    var checkCodePoint = (n - remainder) % n;

    return CharacterFromCodePoint(checkCodePoint);
}

function ValidateCheckCharacter(input) {

    var factor = 1;
    var sum = 0;
    var n = NumberOfValidInputCharacters();
    input = input.toUpperCase();

    // Starting from the right, work leftwards
    // Now, the initial "factor" will always be "1" 
    // since the last character is the check character
    for (var i = input.length - 1; i >= 0; i--) {
        var codePoint = CodePointFromCharacter(input[i]);
        if( codePoint < 0) {
            return false;
        }

        var addend = factor * codePoint;

        // Alternate the "factor" that each "codePoint" is multiplied by
        factor = (factor == 2) ? 1 : 2;

        // Sum the digits of the "addend" as expressed in base "n"
        addend = Math.floor(addend / n) + (addend % n);
        sum += addend;
    }

    var remainder = sum % n;

    return (remainder == 0);
}

// quick test:
console.log ("check character for 'abcde234': %s",
     GenerateCheckCharacter("abcde234"));
console.log( "validate  'abcde2349' : %s " ,
     ValidateCheckCharacter( "abcde2349"));
console.log( "validate  'abcde234X' : %s" ,
     ValidateCheckCharacter( "abcde234X"));