Develop Reference

Any particular reason why this switch-case statement in js is returning undefined?

Here is my code. I have cross checked with online docs and couldn't find any reason for this to not work.
let marks = 90;
switch (marks) {
case 1:
if (marks <= 100 && marks >= 80) {
console.log("Very Good");
}
break;
case 2:
if (marks >= 60 && marks <= 79) {
console.log("Good");
}
break;
case 3:
if (marks >= 30 && marks <= 59) {
console.log("Can do better");
}
break;
case 4:
if (marks < 30) {
console.log("Fail");
}
break;
}

let marks = 90;
switch (marks<=100) {
case marks <= 100 && marks>=80:
console.log("Very Good");
break
case ( marks >=60 && marks <= 79) :
console.log("Good")
break;
case (marks >= 30 && marks <=59):
console.log("Can do better");
break;
case (marks < 30) :
console.log("Fail");
break;
}
This should work your switch cases don't make sense

Switch...case in JS

This simple problem gives me an error. Does not get the correct answer. I will be glad if you help.
let point = 90;
switch (point) {
case point >= 51 && point <= 60:
console.log('Your price: E');
break;
case point >= 61 && point <= 70:
console.log('Your price: D');
break;
case point >= 71 && point <= 80:
console.log('Your price: C');
break;
case point >= 81 && point <= 90:
console.log('Your price: B');
break;
case point >= 91 && point <= 100:
console.log('Your price: A');
break;
default:
console.log('You did not pass');
}
Output: You did not pass

this way
let point = 90;
switch (true) {
case point >= 51 && point <= 60:
console.log('Your price: E');
break;
case point >= 61 && point <= 70:
console.log('Your price: D');
break;
case point >= 71 && point <= 80:
console.log('Your price: C');
break;
case point >= 81 && point <= 90:
console.log('Your price: B');
break;
case point >= 91 && point <= 100:
console.log('Your price: A');
break;
default:
console.log('You did not pass');
}
can you explain why we write true? – Hussein Nadjafli (PO)
The JS switch only works on strict equality.
switch (A) {
case ‘x1’: ...
case ‘x2’: ...
is equivalent to
if (A === ’x1’) { ...
else if (A === ’x2’) { ...
in your code you replace the possible values [’x1’,’x2’,...] with an evaluation like
(point >= 61 && point <= 70)
which returns either true or false
so your code becomes:
if (A === (point >= 51 && point <= 60)) { ...
else if (A === (point >= 61 && point <= 70)) { ...
by replacing the A by true you therefore have a comparison between:
if (true === (point >= 51 && point <= 60)) { ...
else if (true === (point >= 61 && point <= 70)) { ...
You can also do:
function codePrice(val)
{
let code = 'ABCDE'[10 - Math.ceil(val / 10)]
return (!!code) ? `Your price: ${code}` :'You did not pass'
}
console.log( codePrice(90) )

Can also be simplified:
let point = 90;
switch (true) {
case point < 51:
console.log('You did not pass');
break;
case point < 61:
console.log('Your price: E');
break;
case point < 71:
console.log('Your price: D');
break;
case point < 81:
console.log('Your price: C');
break;
case point < 91:
console.log('Your price: B');
break;
default:
console.log('Your price: A');
break
}

How to code else if in short if statement? [duplicate]

This question already has answers here:
shortHand if else-if and else statement
(3 answers)
Closed 3 years ago.
For example, I have code with short if statement and how can I code else if in this code?
var age = 16;
age > 18 ? console.log("> 18") : console.log("< 18");
That would work like this code
var age = 16;
if (age > 18){
console.log("> 18");
}else if(age == 18){
console.log("= 18");
}else{
console.log("< 18");
}

A hybrid approach is recommended in this case:
var age = 16;
if (age == 18) {
console.log("= 18");
} else {
console.log(age > 18 ? "> 18" : "< 18");
}

Do it with ternary expression,
var age = 16;
console.log(age === 18 ? "=18" : age < 18 ? "< 18" : "> 18")
var age = 19;
console.log(age === 18 ? "=18" : age < 18 ? "< 18" : "> 18")
var age = 18;
console.log(age === 18 ? "=18" : age < 18 ? "< 18" : "> 18")

The if way could also be written like this:
var age = 16;
if (age > 18) {
console.log('> 18')
} else if (age == 18) {
console.log("= 18");
} else {
console.log("< 18");
}
So, you could use a nested ternary in the style of the code above:
var age = 16;
age > 18 ? console.log('> 18') : age === 18 ? console.log("18") : console.log("< 18")
Note, it would be more efficient to write the ternary inside console.log, since you are always returning similar values (in this case, strings):
var age = 16;
console.log(age > 18 ? age === 18 ? '18' : '< 18' : '< 18')

What is the closest equivalent to Go's switch without condition in JavaScript?

In Go, I can use a switch without a condition, and instead provide the conditions in the case branches, such as:
package main
import (
"fmt"
"time"
)
func main() {
t := time.Now()
switch {
case t.Hour() < 12:
fmt.Println("Good morning!")
case t.Hour() < 17:
fmt.Println("Good afternoon.")
default:
fmt.Println("Good evening.")
}
}
(Taken from https://tour.golang.org/flowcontrol/11)
What I like about this approach is that it is much cleaner than if-else if-else if-…. Unfortunately, this construct is not possible in JavaScript.
How could I create something that looks like this as closely as possible, using some (weird) language constructs?

You can use virtually the same construct as in Go:
var now = new Date();
switch (true) {
case now.getHours() < 12:
console.log('Good morning');
break;
case now.getHours() < 17:
console.log('Good afternoon');
break;
default:
console.log('Good evening');
}

You could use conditions at the case clause.
var a = 2;
switch (true) { // strict comparison!
case a < 3:
console.log(a + ' is smaller than 3');
break;
}

abusing the language,
var now = new Date();
now.getHours() < 12 && console.log('Good morning') ||
now.getHours() < 17 && console.log('Good afternoon') ||
now.getHours() >= 17 && console.log('Good evening')

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

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:
I've set up a for loop which loops through the 12 days.
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>
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) {
}
}
}

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

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))
);

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"

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]]

You can use the moment libraries local data functions.
Code:
moment.localeData().ordinal(1)
//1st

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";
}

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'

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

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

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.

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

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;
};

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)

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

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

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;
}
}

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')
}
}

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)}`;
}

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]);
}

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

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

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

<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