前几天我写了这个简单的函数。虽然对于一个日期你不需要更大的数字，但这也可以满足更高的数值(1013th, 36021等等……)

var fGetSuffix = function(nPos){

    var sSuffix = "";

    switch (nPos % 10){
        case 1:
            sSuffix = (nPos % 100 === 11) ? "th" : "st";
            break;
        case 2:
            sSuffix = (nPos % 100 === 12) ? "th" : "nd";
            break;
        case 3:
            sSuffix = (nPos % 100 === 13) ? "th" : "rd";
            break;
        default:
            sSuffix = "th";
            break;
    }

    return sSuffix;
};

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

通过将数字分成一个数组并反转，我们可以很容易地使用数组[0]和数组[1]检查数字的最后两位数字。

如果一个数字是十位数数组[1]= 1，它需要“th”。

function getDaySuffix(num)
{
    var array = ("" + num).split("").reverse(); // E.g. 123 = array("3","2","1")
    
    if (array[1] != "1") { // Number is not in the teens
        switch (array[0]) {
            case "1": return "st";
            case "2": return "nd";
            case "3": return "rd";
        }
    }
    
    return "th";
}

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;
     }
}

Intl。PluralRules，标准方法。

我只是想在这里放弃做这个的规范方法，因为似乎没有人知道它。不要白费力气。

如果你想让你的代码

自我记录 易于本地化 用现代标准

-这才是正确的方法。

const english_ordinal_rules = new Intl。PluralRules("en"， {type: "ordinal"}); Const后缀= { 一:“圣”, 二:“和”, 一些:“路”, 其他:“th” }; 函数序数(number/*: number */) { Const category = english_ordinal_rules.select(number); Const suffix =后缀[类别]; 返回(数字+后缀); } // ->字符串 const test =数组(201) .fill () .map((_， index) => index - 100) . map(顺序) . join (" "); console.log(测试);

Intl。PluralRules构造函数(ECMA-402草案) 统一码的六个多元类别

Code-golf

虽然我不建议用你的代码打高尔夫，杀死可读性，但我为那些高尔夫球手想出了一个(92字节):

n=>n+{e:"st",o:"nd",w:"rd",h:"th"}[new Intl.PluralRules("en",{type:"ordinal"}).select(n)[2]]

我强烈推荐这本书，它超级简单易懂。希望有帮助?

它避免使用负整数，即小于1的数字并返回false 如果输入为0，则返回0

function numberToOrdinal(n) {

  let result;

  if(n < 0){
    return false;
  }else if(n === 0){
    result = "0";
  }else if(n > 0){

    let nToString = n.toString();
    let lastStringIndex = nToString.length-1;
    let lastStringElement = nToString[lastStringIndex];

    if( lastStringElement == "1" && n % 100 !== 11 ){
      result = nToString + "st";
    }else if( lastStringElement == "2" && n % 100 !== 12 ){
      result = nToString + "nd";
    }else if( lastStringElement == "3" && n % 100 !== 13 ){
      result = nToString + "rd";
    }else{
      result = nToString + "th";
    }

  }

  return result;
}

console.log(numberToOrdinal(-111));
console.log(numberToOrdinal(0));
console.log(numberToOrdinal(11));
console.log(numberToOrdinal(15));
console.log(numberToOrdinal(21));
console.log(numberToOrdinal(32));
console.log(numberToOrdinal(43));
console.log(numberToOrdinal(70));
console.log(numberToOrdinal(111));
console.log(numberToOrdinal(300));
console.log(numberToOrdinal(101));

输出

false
0
11th
15th
21st
32nd
43rd
70th
111th
300th
101st