将数字转换为字母

是的，有Number#toString(36)和调整。

var value = 10;
document.write((value + 9).toString(36).toUpperCase());

你可以简单地做到这一点，而无需使用数组String.fromCharCode(code)因为字母有连续的代码。例如：String.fromCharCode(1+64)给你"A"，String.fromCharCode(2+64)给你"B"，依此类推。

下面的代码片段将字母表中的字符转换为数字系统

1 = A
2 = B
...
26 = Z
27 = AA
28 = AB
...
78 = BZ
79 = CA
80 = CB

var alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
var result = ""
function printToLetter(number){
    var charIndex = number % alphabet.length
    var quotient = number/alphabet.length
    if(charIndex-1 == -1){
        charIndex = alphabet.length
        quotient--;
    }
    result =  alphabet.charAt(charIndex-1) + result;
    if(quotient>=1){
        printToLetter(parseInt(quotient));
    }else{
        console.log(result)
        result = ""
    }
}

我创建了这个函数是为了在打印时保存字符，但不得不废弃它，因为我不想处理最终可能形成的不正确的单词

只需将字母索引从 0 （A） 增加到 25 （Z）

const letterIndex = 0
const letter = String.fromCharCode(letterIndex + 'A'.charCodeAt(0))
console.log(letter)

更新（5/2/22）：在第二个项目中需要这段代码后，我决定增强以下答案并将其转换为一个名为 alphanumeric-encoder 的现成 NPM 库。如果您不想为此问题构建自己的解决方案，请查看库！

我构建了以下解决方案，作为对@esantos答案的增强。

第一个函数定义有效的查找编码字典。在这里，我使用了英文字母表的所有 26 个字母，但以下内容也可以正常工作："ABCDEFG"、"ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"、"GFEDCBA" 。使用这些词典之一将导致将您的基数 10 转换为具有适当编码数字的基数dictionary.length数。唯一的限制是字典中的每个字符都必须是唯一的。

function getDictionary() {
    return validateDictionary("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
    function validateDictionary(dictionary) {
        for (let i = 0; i < dictionary.length; i++) {
            if(dictionary.indexOf(dictionary[i]) !== dictionary.lastIndexOf(dictionary[i])) {
                console.log('Error: The dictionary in use has at least one repeating symbol:', dictionary[i])
                return undefined
            }
        }
        return dictionary
    }
}

我们现在可以使用这个字典来编码我们的基数 10。

function numberToEncodedLetter(number) {
    //Takes any number and converts it into a base (dictionary length) letter combo. 0 corresponds to an empty string.
    //It converts any numerical entry into a positive integer.
    if (isNaN(number)) {return undefined}
    number = Math.abs(Math.floor(number))
    const dictionary = getDictionary()
    let index = number % dictionary.length
    let quotient = number / dictionary.length
    let result
    
    if (number <= dictionary.length) {return numToLetter(number)}  //Number is within single digit bounds of our encoding letter alphabet
    if (quotient >= 1) {
        //This number was bigger than our dictionary, recursively perform this function until we're done
        if (index === 0) {quotient--}   //Accounts for the edge case of the last letter in the dictionary string
        result = numberToEncodedLetter(quotient)
    }
    if (index === 0) {index = dictionary.length}   //Accounts for the edge case of the final letter; avoids getting an empty string
    
    return result + numToLetter(index)
    function numToLetter(number) {
        //Takes a letter between 0 and max letter length and returns the corresponding letter
        if (number > dictionary.length || number < 0) {return undefined}
        if (number === 0) {
            return ''
        } else {
            return dictionary.slice(number - 1, number)
        }
    }
}

一组编码的字母很棒，但是如果我不能将其转换回以 10 为基数的数字，这对计算机来说就毫无用处。

function encodedLetterToNumber(encoded) {
    //Takes any number encoded with the provided encode dictionary 
    const dictionary = getDictionary()
    let result = 0
    let index = 0
    for (let i = 1; i <= encoded.length; i++) {
        index = dictionary.search(encoded.slice(i - 1, i)) + 1
        if (index === 0) {return undefined} //Attempted to find a letter that wasn't encoded in the dictionary
        result = result + index * Math.pow(dictionary.length, (encoded.length - i))
    }
    return result
}

现在测试一下：

console.log(numberToEncodedLetter(4))     //D
console.log(numberToEncodedLetter(52))    //AZ
console.log(encodedLetterToNumber("BZ"))  //78
console.log(encodedLetterToNumber("AAC")) //705

更新

您还可以使用此函数采用您拥有的短名称格式，并将其返回到基于索引的格式。

function shortNameToIndex(shortName) {
    //Takes the short name (e.g. F6, AA47) and converts to base indecies ({6, 6}, {27, 47})
    if (shortName.length < 2) {return undefined}    //Must be at least one letter and one number
    if (!isNaN(shortName.slice(0, 1))) {return undefined}  //If first character isn't a letter, it's incorrectly formatted
    let letterPart = ''
    let numberPart= ''
    let splitComplete = false
    let index = 1
    do {
        const character = shortName.slice(index - 1, index)
        if (!isNaN(character)) {splitComplete = true}
        if (splitComplete && isNaN(character)) {
            //More letters existed after the numbers. Invalid formatting.
            return undefined    
        } else if (splitComplete && !isNaN(character)) {
            //Number part
            numberPart = numberPart.concat(character)
        } else {
            //Letter part
            letterPart = letterPart.concat(character)
        }
        index++
    } while (index <= shortName.length)
    numberPart = parseInt(numberPart)
    letterPart = encodedLetterToNumber(letterPart)
    return {xIndex: numberPart, yIndex: letterPart}
}

有一个单行函数numAbbr用于将数字转换为字符串，例如

1 => A
2 => B
...
26 => Z
27 => AA
28 => AB
...
702 => ZZ
703 => AAA

const numAbbr = num => num <= 0 ? '' : numAbbr(Math.floor((num - 1) / 26)) + String.fromCharCode((num - 1) % 26 + 65);

和相反的函数abbrNum将AAA等字符串转换为数字

const abbrNum = abbr => abbr.toUpperCase().split("").reduce((acc, val) => acc * 26 + val.charCodeAt(0) - 64, 0);

的代码答案真的过于复杂，因为它可以通过一个简单的循环来实现

function colToLetter(number){
  let result = '';
  // number = number - 1; // If starting from 1
  do {
    const letter = String.fromCharCode(65 + (number % 26));
    result = letter + result;
    number = Math.floor(number / 26) - 1;
  } while (number >= 0)
  return result;
}
console.log(colToLetter(0));
console.log(colToLetter(25));
console.log(colToLetter(26));
console.log(colToLetter(702));
console.log(colToLetter(728));

这提供了类似 excel 的编号

这可以帮助你

static readonly string[] Columns_Lettre = new[] { "A", "B", "C"};
public static string IndexToColumn(int index)
    {
        if (index <= 0)
            throw new IndexOutOfRangeException("index must be a positive number");
        if (index < 4)
            return Columns_Lettre[index - 1];
        else
            return index.ToString();
    }