Question

我正在尝试制作一个playfair密码方生成器。

如果你知道那是什么，那么你就知道你有一把钥匙然后进入一个正方形，其中不在键中的其余字母按字母顺序输入（也删除了J键）。 / p>

所以目前我有这个功能：

function square(usrInputKey){
    var alpha = ["A","B","C","D","E","F","G","H","I","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"];
    var key = usrInputKey.toUpperCase()
    keyword = key.split('')
    var keysquare=keyword
    count=keyword.length
    for (var p=0;p<alpha.length;p++) {
        //document.write(" Alpha: " + alpha[p] + "<br>")
        if ( keyword[0] != alpha[p] && keyword[1] != alpha[p] && keyword[2] != alpha[p] ) {
            keysquare[count]=alpha[p]
            count++
        }
    }
            return keysquare
}

该函数具有关键字的输入，该关键字是用于测试perposes的单词键。

此功能有效，输出为：K，E，Y，A，B，C，D，F，G，H，I，L，M，N，O，P，Q，R，S，T， U，V，W，X，Z

但是只有在关键字长度为3个字符时才有效。如果它使用任何数量的关键字而不是硬编码3或者具有不同可能长度的语句，我希望它。

我尝试过.every（），但是只检查一个数组字符，然后输出密钥，然后输出完整的字母，包括密钥中的字母。

任何人都可以解决这个问题吗？还是必须硬编码到某个密钥长度？

Answer 1

这是一个生成所需表格的函数，以及一些茉莉花测试，只是因为：

describe('Key generator test', function () {

    var wordControl = {};
    var alphabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',  'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'];

    beforeEach(function() {
        alphabet.forEach(function(letter) {
            wordControl[letter] = 0;
        });
    });

    function square(key) {
        var result = [];
        key = key.toUpperCase();
        key = key.replace('J', 'I');
        var keyArray = key.split('');
        keyArray.forEach(function(letter) {
            if(wordControl[letter] == 0) {
                result.push(letter);
                wordControl[letter] = 1;
            }
        });

        alphabet.forEach(function(letter) {
            if(wordControl[letter] == 0) {
                result.push(letter);
            }
        });

        return result;
    }

    it('should generate a table starting with tke key followed by the sorted alphabet', function() {
        var key = 'KAWABANGA';
        var generatedTable = square(key);
        expect(generatedTable.join('')).toEqual('KAWBNGCDEFHILMOPQRSTUVXYZ');
    });

    it('should replace J with I', function() {
        var key = 'JONAS';
        var generatedTable = square(key);
        expect(generatedTable.join('')).toEqual('IONASBCDEFGHKLMPQRTUVWXYZ');
    });

    it('should not fail due to key being more than 25 characters', function() {
        var key = 'QWERTYUIOPASDFGHJKLZXCVBNMMNBVCXZLKJHGFDSAPOIUYTREWQ';
        var generatedTable = square(key);
        expect(generatedTable.join('')).toEqual('QWERTYUIOPASDFGHKLZXCVBNM');
    });
});

我猜你仍然需要在5x5矩阵中转换结果数组，以便继续使用playfair密码算法。如果你遇到麻烦，请告诉我。

Answer 2

我认为这是一个有效的例子。 join和console.log是为了我的方便。

function square(usrInputKey){
  var alpha =  ["A","B","C","D","E","F","G","H","I","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"];
  var key = usrInputKey.toUpperCase();
  keyword = key.split('');
  var newAlpha = [];
  for (var k = 0; k < alpha.length; k++) {
    found = false;
    for (var j = 0; j < key.length; j++) {
      if (alpha[k].indexOf(key[j]) > -1){
        found = true;
      }
    }
    if (!found){
      newAlpha.push(alpha[k]);
    }
  }
  newArray = keyword.concat(newAlpha);
  return newArray.join(',');
}
console.log(square('KEYWORD'));

另外，在适当情况下使用分号是很好的主意（行尾等）

如何在javascript中将数组值与另一组数组值进行比较

2 个答案: