Add st, nd, rd and th (ordinal) suffix to a number

Question

I would like to dynamically generate a string of text based on a current day. So, for example, if it is day 1 then I would like my code to generate = "Its the <dynamic>1*<dynamic string>st</dynamic string>*</dynamic>".

There are 12 days in total so I have done the following:

1. I've set up a for loop which loops through the 12 days.

2. In my html I have given my element a unique id with which to target it, see below:

   <h1 id="dynamicTitle" class="CustomFont leftHeading shadow">On The <span></span> <em>of rest of generic text</em></h1>
   

3. Then, inside my for loop I have the following code:

   $("#dynamicTitle span").html(i);
   var day = i;
   if (day == 1) {
       day = i + "st";
   } else if (day == 2) {
       day = i + "nd"
   } else if (day == 3) {
       day = i + "rd"
   }
   

UPDATE

This is the entire for loop as requested:

$(document).ready(function () {
    for (i = 1; i <= 12; i++) {
        var classy = "";
        if (daysTilDate(i + 19) > 0) {
            classy = "future";
            $("#Day" + i).addClass(classy);
            $("#mainHeading").html("");
            $("#title").html("");
            $("#description").html("");
        } else if (daysTilDate(i + 19) < 0) {
            classy = "past";
            $("#Day" + i).addClass(classy);
            $("#title").html("");
            $("#description").html("");
            $("#mainHeading").html("");
            $(".cta").css('display', 'none');
            $("#Day" + i + " .prizeLink").attr("href", "" + i + ".html");
        } else {
            classy = "current";
            $("#Day" + i).addClass(classy);
            $("#title").html(headings[i - 1]);
            $("#description").html(descriptions[i - 1]);
            $(".cta").css('display', 'block');
            $("#dynamicImage").attr("src", ".." + i + ".jpg");
            $("#mainHeading").html("");
            $(".claimPrize").attr("href", "" + i + ".html");
            $("#dynamicTitle span").html(i);
            var day = i;
            if (day == 1) {
                day = i + "st";
            } else if (day == 2) {
                day = i + "nd"
            } else if (day == 3) {
                day = i + "rd"
            } else if (day) {
            }
        }
    }

Answer 1

The rules are as follows:

• st is used with numbers ending in 1 (e.g. 1st, pronounced first)
• nd is used with numbers ending in 2 (e.g. 92nd, pronounced ninety-second)
• rd is used with numbers ending in 3 (e.g. 33rd, pronounced thirty-third)
• As an exception to the above rules, all the "teen" numbers ending with 11, 12 or 13 use -th (e.g. 11th, pronounced eleventh, 112th, pronounced one hundred [and] twelfth)
• th is used for all other numbers (e.g. 9th, pronounced ninth).

The following JavaScript code (rewritten in Jun '14) accomplishes this:

function ordinal_suffix_of(i) {
    var j = i % 10,
        k = i % 100;
    if (j == 1 && k != 11) {
        return i + "st";
    }
    if (j == 2 && k != 12) {
        return i + "nd";
    }
    if (j == 3 && k != 13) {
        return i + "rd";
    }
    return i + "th";
}

Sample output for numbers between 0-115:

  0  0th
  1  1st
  2  2nd
  3  3rd
  4  4th
  5  5th
  6  6th
  7  7th
  8  8th
  9  9th
 10  10th
 11  11th
 12  12th
 13  13th
 14  14th
 15  15th
 16  16th
 17  17th
 18  18th
 19  19th
 20  20th
 21  21st
 22  22nd
 23  23rd
 24  24th
 25  25th
 26  26th
 27  27th
 28  28th
 29  29th
 30  30th
 31  31st
 32  32nd
 33  33rd
 34  34th
 35  35th
 36  36th
 37  37th
 38  38th
 39  39th
 40  40th
 41  41st
 42  42nd
 43  43rd
 44  44th
 45  45th
 46  46th
 47  47th
 48  48th
 49  49th
 50  50th
 51  51st
 52  52nd
 53  53rd
 54  54th
 55  55th
 56  56th
 57  57th
 58  58th
 59  59th
 60  60th
 61  61st
 62  62nd
 63  63rd
 64  64th
 65  65th
 66  66th
 67  67th
 68  68th
 69  69th
 70  70th
 71  71st
 72  72nd
 73  73rd
 74  74th
 75  75th
 76  76th
 77  77th
 78  78th
 79  79th
 80  80th
 81  81st
 82  82nd
 83  83rd
 84  84th
 85  85th
 86  86th
 87  87th
 88  88th
 89  89th
 90  90th
 91  91st
 92  92nd
 93  93rd
 94  94th
 95  95th
 96  96th
 97  97th
 98  98th
 99  99th
100  100th
101  101st
102  102nd
103  103rd
104  104th
105  105th
106  106th
107  107th
108  108th
109  109th
110  110th
111  111th
112  112th
113  113th
114  114th
115  115th

Answer 2

From Shopify

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

[-4,-1,0,1,2,3,4,10,11,12,13,14,20,21,22,100,101,111].forEach(
  n => console.log(n + ' -> ' + getNumberWithOrdinal(n))
);

Answer 3

Minimal one-line approach for ordinal suffixes

function nth(n){return["st","nd","rd"][((n+90)%100-10)%10-1]||"th"}

(this is for positive integers, see below for other variations)

Explanation

Start with an array with the suffixes ["st", "nd", "rd"]. We want to map integers ending in 1, 2, 3 (but not ending in 11, 12, 13) to the indexes 0, 1, 2.

Other integers (including those ending in 11, 12, 13) can be mapped to anything else—indexes not found in the array will evaluate to undefined. This is falsy in javascript and with the use of logical or (|| "th") the expression will return "th" for these integers, which is exactly what we want.

The expression ((n + 90) % 100 - 10) % 10 - 1 does the mapping. Breaking it down:

• (n + 90) % 100: This expression takes the input integer − 10 mod 100, mapping 10 to 0, ... 99 to 89, 0 to 90, ..., 9 to 99. Now the integers ending in 11, 12, 13 are at the lower end (mapped to 1, 2, 3).
• - 10: Now 10 is mapped to −10, 19 to −1, 99 to 79, 0 to 80, ... 9 to 89. The integers ending in 11, 12, 13 are mapped to negative integers (−9, −8, −7).
• % 10: Now all integers ending in 1, 2, or 3 are mapped to 1, 2, 3. All other integers are mapped to something else (11, 12, 13 are still mapped to −9, −8, −7).
• - 1: Subtracting one gives the final mapping of 1, 2, 3 to 0, 1, 2.

Verifying that it works

function nth(n){return["st","nd","rd"][((n+90)%100-10)%10-1]||"th"}

//test integers from 1 to 124
for(var r = [], i = 1; i < 125; i++) r.push(i + nth(i));

//output result
document.getElementById('result').innerHTML = r.join('<br>');

<div id="result"></div>

Variations

Allowing negative integers:

function nth(n){return["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"}

function nth(n){return["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"}

//test integers from 15 to -124
for(var r = [], i = 15; i > -125; i--) r.push(i + nth(i));

//output result
document.getElementById('result').innerHTML = r.join('<br>');

<div id="result"></div>

In ES6 fat arrow syntax (anonymous function):

n=>["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"

Answer 4

Intl.PluralRules, the standard method.

I would just like to drop the canonical way of doing this in here, as nobody seems to know it. Do not reinvent the wheel.

If you want your code to be

• self-documenting
• easy to localize
• with the modern standard

― this is the way to go.

const english_ordinal_rules = new Intl.PluralRules("en", {type: "ordinal"});
const suffixes = {
    one: "st",
    two: "nd",
    few: "rd",
    other: "th"
};
function ordinal(number/*: number */) {
    const category = english_ordinal_rules.select(number);
    const suffix = suffixes[category];
    return (number + suffix);
} // -> string

const test = Array(201)
    .fill()
    .map((_, index) => index - 100)
    .map(ordinal)
    .join(" ");
console.log(test);

• The Intl.PluralRules constructor (Draft ECMA-402)
• Unicode’s six plurality categories

---

Code-golf

While I do not recommend golfing with your code and killing the readability, I came up with one for those golfers (92 bytes):

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

Answer 5

You can use the moment libraries local data functions.

Code:

moment.localeData().ordinal(1)
//1st

Answer 6

By splitting the number into an array and reversing we can easily check the last 2 digits of the number using array[0] and array[1].

If a number is in the teens array[1] = 1 it requires "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";
}

Answer 7

You've only got 12 days? I'd be tempted to make it just a simple lookup array:

var suffixes = ['','st','nd','rd','th','th','th','th','th','th','th','th','th'];

then

var i = 2;
var day = i + suffixes[i]; // result: '2nd'

or

var i = 8;
var day = i + suffixes[i]; // result: '8th'

Answer 8

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

Answer 9

I wrote this function to solve this problem:

// 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';
    }
};

With this you can just put .addSuffix() to any number and it will result in what you want. For example:

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

Answer 10

An alternative version of the ordinal function could be as follows:

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";
}

The variables are named more explicitly, uses camel case convention, and might be faster.

Answer 11

const getOrdinalNum = (n) => n + (n > 0 ? ['th', 'st', 'nd', 'rd'][(n > 3 && n < 21) || n % 10 > 3 ? 0 : n % 10] : '');

Answer 12

I wrote this simple function the other day. Although for a date you don't need the larger numbers, this will cater for higher values too (1013th, 36021st etc...)

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;
};

Answer 13

function ordsfx(a){return["th","st","nd","rd"][(a=~~(a<0?-a:a)%100)>10&&a<14||(a%=10)>3?0:a]}

See annotated version at https://gist.github.com/furf/986113#file-annotated-js

Short, sweet, and efficient, just like utility functions should be. Works with any signed/unsigned integer/float. (Even though I can't imagine a need to ordinalize floats)

Answer 14

Strongly recommend the excellent date-fns library. Fast, modular, immutable, works with standard dates.

import * as DateFns from 'date-fns';

const ordinalInt = DateFns.format(someInt, 'do');

See date-fns docs: https://date-fns.org/v2.0.0-alpha.9/docs/format

Answer 15

A version of the Intl.PluralRules answer that TypeScript type checking approves of:

const englishOrdinalRules = new Intl.PluralRules('en', { type: 'ordinal' })
/**
 * Format a number as an ordinal, e.g. `3` to `3rd`.
 * @link https://stackoverflow.com/a/57518703/223225
 * @param {number} number To format
 */
export function formatNumberOrdinal(number) {
    const category = englishOrdinalRules.select(number)

    switch (category) {
        case 'one': {
            return `${number}st`
        }

        case 'two': {
            return `${number}nd`
        }

        case 'few': {
            return `${number}rd`
        }

        default: {
            return `${number}th`
        }
    }
}

Answer 16

Here is another option.

function getOrdinalSuffix(day) {
        
    if(/^[2-3]?1$/.test(day)){
     return 'st';
    } else if(/^[2-3]?2$/.test(day)){
     return 'nd';
    } else if(/^[2-3]?3$/.test(day)){
     return 'rd';
    } else {
     return 'th';
    }
        
}
    
console.log(getOrdinalSuffix('1'));
console.log(getOrdinalSuffix('13'));
console.log(getOrdinalSuffix('22'));
console.log(getOrdinalSuffix('33'));

Notice the exception for the teens? Teens are so akward!

Edit: Forgot about 11th and 12th

Answer 17

Old one I made for my stuff...

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;
     }
}

Answer 18

I wrote this function for higher numbers and all test cases

function numberToOrdinal(num) {
    if (num === 0) {
        return '0'
    };
    let i = num.toString(), j = i.slice(i.length - 2), k = i.slice(i.length - 1);
    if (j >= 10 && j <= 20) {
        return (i + 'th')
    } else if (j > 20 && j < 100) {
        if (k == 1) {
            return (i + 'st')
        } else if (k == 2) {
            return (i + 'nd')
        } else if (k == 3) {
            return (i + 'rd')
        } else {
            return (i + 'th')
        }
    } else if (j == 1) {
        return (i + 'st')
    } else if (j == 2) {
        return (i + 'nd')
    } else if (j == 3) {
        return (i + 'rd')
    } else {
        return (i + 'th')
    }
}

Answer 19

Here's a slightly different approach (I don't think the other answers do this). I'm not sure whether I love it or hate it, but it works!

export function addDaySuffix(day: number) { 
  const suffixes =
      '  stndrdthththththththththththththththththstndrdthththththththst';
    const startIndex = day * 2;

    return `${day}${suffixes.substring(startIndex, startIndex + 2)}`;
  }

Answer 20

I would like to quote the answer available in the link

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

Answer 21

I wanted to provide a functional answer to this question to complement the existing answer:

const ordinalSuffix = ['st', 'nd', 'rd']
const addSuffix = n => n + (ordinalSuffix[(n - 1) % 10] || 'th')
const numberToOrdinal = n => `${n}`.match(/1\d$/) ? n + 'th' : addSuffix(n)

we've created an array of the special values, the important thing to remember is arrays have a zero based index so ordinalSuffix[0] is equal to 'st'.

Our function numberToOrdinal checks if the number ends in a teen number in which case append the number with 'th' as all then numbers ordinals are 'th'. In the event that the number is not a teen we pass the number to addSuffix which adds the number to the ordinal which is determined by if the number minus 1 (because we're using a zero based index) mod 10 has a remainder of 2 or less it's taken from the array, otherwise it's 'th'.

sample output:

numberToOrdinal(1) // 1st
numberToOrdinal(2) // 2nd
numberToOrdinal(3) // 3rd
numberToOrdinal(4) // 4th
numberToOrdinal(5) // 5th
numberToOrdinal(6) // 6th
numberToOrdinal(7) // 7th
numberToOrdinal(8) // 8th
numberToOrdinal(9) // 9th
numberToOrdinal(10) // 10th
numberToOrdinal(11) // 11th
numberToOrdinal(12) // 12th
numberToOrdinal(13) // 13th
numberToOrdinal(14) // 14th
numberToOrdinal(101) // 101st

Answer 22

I strongly recommend this, it is super easy and straightforward to read. I hope it help?

• It avoid the use of negative integer i.e number less than 1 and return false
• It return 0 if input is 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));

OUTPUT

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

Answer 23

This is for one liners and lovers of es6

let i= new Date().getDate 

// I can be any number, for future sake we'll use 9

const j = I % 10;
const k = I % 100;

i = `${i}${j === 1 &&  k !== 11 ? 'st' : j === 2 && k !== 12 ? 'nd' : j === 3 && k !== 13 ? 'rd' : 'th'}`}

console.log(i) //9th

Another option for +be number would be:

console.log(["st","nd","rd"][((i+90)%100-10)%10-1]||"th"]

Also to get rid of the ordinal prefix just use these:

console.log(i.parseInt("8th"))

console.log(i.parseFloat("8th"))

feel free to modify to suit you need

Answer 24

<p>31<sup>st</sup> March 2015</p>

You can use

1<sup>st</sup> 2<sup>nd</sup> 3<sup>rd</sup> 4<sup>th</sup>

for positioning the suffix