Convert integer into its character equivalent, where 0 => a, 1 => b, etc

问题

I want to convert an integer into its character equivalent based on the alphabet. For example:

0 => a
1 => b
2 => c
3 => d

etc. I could build an array and just look it up when I need it but I’m wondering if there’s a built in function to do this for me. All the examples I’ve found via Google are working with ASCII values and not a character’s position in the alphabet.

回答1:

Assuming you want lower case letters:

var chr = String.fromCharCode(97 + n); // where n is 0, 1, 2 ...

97 is the ASCII code for lower case 'a'. If you want uppercase letters, replace 97 with 65 (uppercase 'A'). Note that if n > 25, you will get out of the range of letters.

回答2:

Will be more portable in case of extending to other alphabets:

char='abcdefghijklmnopqrstuvwxyz'[code]

or, to be more compatible (with our beloved IE):

char='abcdefghijklmnopqrstuvwxyz'.charAt(code);

回答3:

If you don't mind getting multi-character strings back, you can support arbitrary positive indices:

function idOf(i) {
    return (i >= 26 ? idOf((i / 26 >> 0) - 1) : '') +  'abcdefghijklmnopqrstuvwxyz'[i % 26 >> 0];
}

idOf(0) // a
idOf(1) // b
idOf(25) // z
idOf(26) // aa
idOf(27) // ab
idOf(701) // zz
idOf(702) // aaa
idOf(703) // aab

(Not thoroughly tested for precision errors :)

回答4:

A simple answer would be (26 characters):

String.fromCharCode(97+n);

If space is precious you could do the following (20 characters):

(10+n).toString(36);

Think about what you could do with all those extra bytes!

How this works is you convert the number to base 36, so you have the following characters:

0123456789abcdefghijklmnopqrstuvwxyz
^         ^
n        n+10

By offsetting by 10 the characters start at a instead of 0.

Not entirely sure about how fast running the two different examples client-side would compare though.

回答5:

Javascript's String.fromCharCode(code1, code2, ..., codeN) takes an infinite number of arguments and returns a string of letters whose corresponding ASCII values are code1, code2, ... codeN. Since 97 is 'a' in ASCII, we can adjust for your indexing by adding 97 to your index.

function indexToChar(i) {
  return String.fromCharCode(i+97); //97 in ASCII is 'a', so i=0 returns 'a', 
                                    // i=1 returns 'b', etc
}

回答6:

Use String.fromCharCode. This returns a string from a Unicode value, which matches the first 128 characters of ASCII.

var a = String.fromCharCode(97);

回答7:

There you go: (a-zA-Z)

function codeToChar( number ) {
  if ( number >= 0 && number <= 25 ) // a-z
    number = number + 97;
  else if ( number >= 26 && number <= 51 ) // A-Z
    number = number + (65-26);
  else
    return false; // range error
  return String.fromCharCode( number );
}

input: 0-51, or it will return false (range error);

OR:

var codeToChar = function() {
  var abc = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ".split("");
  return function( code ) {
    return abc[code];
  };
})();

returns undefined in case of range error. NOTE: the array will be created only once and because of closure it will be available for the the new codeToChar function. I guess it's even faster then the first method (it's just a lookup basically).

回答8:

I don't like all the solutions that use magic numbers like 97 or 36.

const A = 'A'.charCodeAt(0);

let numberToCharacter = number => String.fromCharCode(A + number);

let characterToNumber = character => character.charCodeAt(0) - A;

this assumes uppercase letters and starts 'A' at 0.

回答9:

The only problemo with @mikemaccana's great solution is that it uses the binary >> operator which is costly, performance-wise. I suggest this modification to his great work as a slight improvement that your colleagues can perhaps read more easily.

const getColumnName = (i) => {
     const previousLetters = (i >= 26 ? getColumnName(Math.floor(i / 26) -1 ) : '');
     const lastLetter = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[i % 26]; 
     return previousLetters + lastLetter;
}

Or as a one-liner

const getColumnName = i => (i >= 26 ? getColumnName(Math.floor(i / 26) -1 ) : '') + 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[i % 26];

Example:

getColumnName(0); // "A"
getColumnName(1); // "B"
getColumnName(25); // "Z"
getColumnName(26); // "AA"
getColumnName(27); // "AB"
getColumnName(80085) // "DNLF"

回答10:

Try

(n+10).toString(36)

chr = n=>(n+10).toString(36);

for(i=0; i<26; i++) console.log(`${i} => ${ chr(i) }`);

回答11:

Assuming you want uppercase case letters:

function numberToLetter(num){
        var alf={
            '0': 'A', '1': 'B', '2': 'C', '3': 'D', '4': 'E', '5': 'F', '6': 'G'
        };
        if(num.length== 1) return alf[num] || ' ';
        return num.split('').map(numberToLetter);
    }

Example:

numberToLetter('023') is ["A", "C", "D"]

numberToLetter('5') is "F"

回答12:

It generates random number and char for phone verification or something else.

function randomIntFromInterval(min,max)
{
    return Math.floor(Math.random()*(max-min+1)+min);
}




function generateRandomVerification(length){
let char;
let sum ="";
  for(let i=0;i < length;i++){
    if(Math.round(Math.random())){
      random = randomIntFromInterval(65,90);
      char = String.fromCharCode(random);//65-90
      sum = sum + char;
      console.log("CHAR: ", char);
    }else{
      random = randomIntFromInterval(48,57);
      char = String.fromCharCode(random);//48-57
      sum = sum + char;
      console.log("CHAR: ", char);
    }
  }
  alert(sum);
}

generateRandomVerification(5);

Here is link

来源：https://stackoverflow.com/questions/3145030/convert-integer-into-its-character-equivalent-where-0-a-1-b-etc

标签

javascript

integer

alphabet