我尝试在 JavaScript 中构建 ROT 13 密码，我做到了，但控制台中未显示输出，请检查其中出了什么问题

Question

一种常见的现代用法是ROT13密码，其中字母的值移动了13位。因此是'A'↔'N'， 'B'↔'O'，以此类推。

function rot13(str) {
    let newStr="";
  let upperAlph=["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","W","X","Y","Z"]
  let rotAlph=["N","O","P","Q","R","S","T","U","V","W","X","Y","Z","A","B","C","D","E","F","G","H","I","J","K","L","M"];
  for(let i=0; i<str.length; i++){
 
    let letter=str[i];
    let letTest=/[A-Z]/g;
    if(letTest.test(letter)){
      let indexOfLetter=upperAlph.indexOf(letter);
      let newLetter=rotAlph[indexOfLetter];
      newStr+newLetter;
    }
    else{
      newStr+letter;
    }
  }
  return newStr;
}
console.log(rot13("SERR PBQR PNZC"));

Answer 1

修复你的代码，看看第12行和第15行，我相信:

function rot13(str) {
    let newStr="";
  let upperAlph=["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","W","X","Y","Z"]
  let rotAlph=["N","O","P","Q","R","S","T","U","V","W","X","Y","Z","A","B","C","D","E","F","G","H","I","J","K","L","M"];
  for(let i=0; i<str.length; i++){
 
    let letter=str[i];
    let letTest=/[A-Z]/g;
    if(letTest.test(letter)){
      let indexOfLetter=upperAlph.indexOf(letter);
      let newLetter=rotAlph[indexOfLetter];
      newStr += newLetter;
    }
    else{
      newStr += letter;
    }
  }
  return newStr;
}
console.log(rot13("SERR PBQR PNZC"));

我也会使用映射类型的对象，而不是两个数组:

alphMap = {A: 'N', B: 'O', C: 'P', ..., Z: 'M'};

你可以像这样找到你的替代字母:

let newLetter = alphMap[letter];

不需要匹配索引

Answer 2

您错过了upperAlph的U和V

    function rot13(str) {
    let newStr="";
    let upperAlph=["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
  let rotAlph=["N","O","P","Q","R","S","T","U","V","W","X","Y","Z","A","B","C","D","E","F","G","H","I","J","K","L","M"];
  for(let i=0; i<str.length; i++){
 
    let letter=str[i];
    let letTest=/[A-Z]/g;
    if(letTest.test(letter)){
      let indexOfLetter=upperAlph.indexOf(letter);
      let newLetter=rotAlph[indexOfLetter];
      newStr += newLetter;
    }
    else{
      newStr += letter;
    }
  }
  return newStr;
}