我写了这个函数来解决这个问题:

// this is for adding the ordinal suffix, turning 1, 2 and 3 into 1st, 2nd and 3rd
Number.prototype.addSuffix=function(){
    var n=this.toString().split('.')[0];
    var lastDigits=n.substring(n.length-2);
    //add exception just for 11, 12 and 13
    if(lastDigits==='11' || lastDigits==='12' || lastDigits==='13'){
        return this+'th';
    }
    switch(n.substring(n.length-1)){
        case '1': return this+'st';
        case '2': return this+'nd';
        case '3': return this+'rd';
        default : return this+'th';
    }
};

有了这个，你可以把. addsuffix()放在任何数字上，它就会得到你想要的结果。例如:

var number=1234;
console.log(number.addSuffix());
// console will show: 1234th

序数函数的另一个版本可以如下所示:

function toCardinal(num) {
    var ones = num % 10;
    var tens = num % 100;

    if (tens < 11 || tens > 13) {
        switch (ones) {
            case 1:
                return num + "st";
            case 2:
                return num + "nd";
            case 3:
                return num + "rd";
        }
    }

    return num + "th";
}

变量的命名更加明确，使用驼峰大小写约定，并且可能更快。

function getSuffix(n) {return n < 11 || n > 13 ? ['st', 'nd', 'rd', 'th'][Math.min((n - 1) % 10, 3)] : 'th'}

函数ordsfx(){返回(“th”、“圣”、“和”,“路”][(= ~ ~(一个< 0 ? a: a) % 100) > 10一个< 14 | | (% = 10)> 3 ?0:]}

参见https://gist.github.com/furf/986113#file-annotated-js的注释版本

就像效用函数应该做的那样，简短、可爱、高效。适用于任何有符号/无符号整数/浮点数。(尽管我无法想象有必要对浮点数进行ordinalize)

以前我为我的东西做的…

function convertToOrdinal(number){
    if (number !=1){
        var numberastext = number.ToString();
        var endchar = numberastext.Substring(numberastext.Length - 1);
        if (number>9){
            var secondfromendchar = numberastext.Substring(numberastext.Length - 1);
            secondfromendchar = numberastext.Remove(numberastext.Length - 1);
        }
        var suffix = "th";
        var digit = int.Parse(endchar);
        switch (digit){
            case 3:
                if(secondfromendchar != "1"){
                    suffix = "rd";
                    break;
                }
            case 2:
                if(secondfromendchar != "1"){
                    suffix = "nd";
                    break;
                }
            case 1:
                if(secondfromendchar != "1"){
                    suffix = "st";
                    break;
                }
            default:
                suffix = "th";
                break;
         }
            return number+suffix+" ";
     } else {
            return;
     }
}

我想引用链接中提供的答案

function ordinal(n) {
  var s = ["th", "st", "nd", "rd"];
  var v = n%100;
  return n + (s[(v-20)%10] || s[v] || s[0]);
}