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Last value of my FizzBuzz function returns undefined

I'm trying to run the FizzBuzz problem solving, it seems to be working, however when it reaches the user input number it returns undefined.
Here is my code:
const prompt = require("prompt-sync")();
let userChoice = parseInt(prompt("Please enter a number: "));
function fizzBuzz () {
for (let i=1 ; i <= userChoice; i++){
if (i % 3 === 0 && i % 5 ===0) {
console.log("FizzBuzz");
}
else if (i % 3 === 0){
console.log("Fizz");
}
else if (i % 5 === 0){
console.log("Buzz");
}
else {
console.log(i);
}
}
}
console.log(fizzBuzz(userChoice));
This is the outcome in the console :

You need to actually return a value from your fizzBuzz() function. At the moment, that function is returning nothing, so when you log its return value, you're getting undefined.
function fizzBuzz (userChoice) {
let output = [ ];
for (let i=1 ; i <= userChoice; i++){
if (i % 3 === 0 && i % 5 ===0) {
output.push("FizzBuzz");
}
else if (i % 3 === 0){
output.push("Fizz");
}
else if (i % 5 === 0){
output.push("Buzz");
}
else {
output.push(i);
}
}
return output.join("\n");
}
Alternatively, just don't wrap your actual function call in a console.log, so your last line will simply be:
fizzBuzz(userChoice);

Fizz Buzz question with extra difficulty layer

I am trying to solve the Fizz Buzz question with an extra layer.
This is what I have so far. All this works fine but I need one more extra condition.
It's JS Unit testing
For any other input (valid or otherwise) return a string representation of the input
printFizzBuzz = function(input) {
// Your code goes here
for (var i = 1; i <= 20; i++) {
if (i % 3 === 0 && i % 5 === 0) {
console.log("FizzBuzz");
} else if (i % 3 === 0) {
console.log("FizzBuzz");
} else if (i % 5 === 0) {
console.log("Buzz");
} else {
console.log(i);
}
}
};
Thank you!

You can use .toString() (JS toString) to change the number in string:
function printFizzBuzz(input) {
// Your code goes here
for (var i = 1; i <= 20; i++) {
if (i % 3 === 0 && i % 5 === 0) {
console.log("FizzBuzz");
} else if (i % 3 === 0) {
console.log("FizzBuzz");
} else if (i % 5 === 0) {
console.log("Buzz");
} else {
console.log(i.toString());
}
}
};
printFizzBuzz();

Looking at the question, it is not asking for a loop, it is asking for any value passed to the function. So the answer should look more like this
function printFizzBuzz(input) {
// Your code goes here
if (typeof input !== ‘number’) {
return String(input)
}
let by3 = input % 3
let by5 = input % 5
switch(0) {
case by3 + by5:
return "FizzBuzz"
case by3:
return "Fizz"
case by5:
return "Buzz"
default:
return input.toString()
}
}
}

If I was the interviewer, I would prefer the answer to look more like this
function printFizzBuzz(input) {
// Your code goes here
for (var i = 1; i <= 20; i++) {
let by3 = i % 3
let by5 = i % 5
switch(0) {
case by3 + by5:
console.log("FizzBuzz")
break
case by3:
console.log("Fizz")
break
case by5:
console.log("Buzz")
break
default:
console.log(i.toString())
}
}
}
}
printFizzBuzz()

FizzBuzz only prints numbers divisible by three

I am learning JavaScript, and an exercise that I am doing...I don't seem to "get" it.
The objective to write a program using console.log that prints all numbers from 1 to 100, with exceptions.
The program should print "FizzBuzz" if the number is divisible by 3 and 5.
The program should print "Fizz" only if the number is divisible by 3.
The program should print "Buzz" only if the number is divisible by 5.
If these exceptions do not apply to the numbers from 1 to 100, the number on its own should be printed.
Here is my code:
for (i = 0; i <= 100; i++) {
if (i % 3 == 0)
if (i % 5 == 0) {
console.log("FizzBuzz")
} else if (i % 3 == 0) {
console.log("Fizz")
} else if (i % 5 == 0) {
console.log("Buzz")
} else {
console.log(i)
}
}
Of course, this code does not work.
The numbers that do not apply to the exceptions do not print. Numbers from 1 to 100 do not print.
Any help explaining why...I would be very thankful.
Thank you.

Your attempt doesn't work, since it only logs those values to the console that are i%3 === 0, since the first if has to be true before the second block is entered.
You can see this if you log the numbers that get printed:
for (i = 0; i <= 100; i++) {
if (i % 3 == 0) // only if this returns "true" the next block will execute
if (i % 5 == 0) {
console.log("FizzBuzz " + i)
} else if (i % 3 == 0) {
console.log("Fizz " + i)
} else if (i % 5 == 0) {
console.log("Buzz " + i)
} else {
console.log(i)
}
}
Combine the first two if statements and it works!
for (i = 0; i <= 100; i++) {
if (i % 3 == 0 && i % 5 == 0) {
console.log("FizzBuzz")
} else if (i % 3 == 0) {
console.log("Fizz")
} else if (i % 5 == 0) {
console.log("Buzz")
} else {
console.log(i)
}
}

You need to combine the first two if statements with && (a boolean operator that means "and". For it to become true, both of the statements must be true. If one of them is false, it becomes false).
JSFiddle (open the console to see it working): http://jsfiddle.net/7236jnx4/
You can not just have this code:
if(i%3==0)
if(i%5==0){
console.log("FizzBuzz");
}
Only numbers that are divisible by 3 will be checked by the other if statements. The first two if statements need to be combined together for it to work:
if(i%5==0&&i%3==0){
console.log("FizzBuzz");
}
for (let i = 0; i <= 100; i++) {
if (i % 5 == 0&&i%3==0) {
console.log("FizzBuzz");
} else if (i % 3 == 0) {
console.log("Fizz")
} else if (i % 5 == 0) {
console.log("Buzz")
} else {
console.log(i);
}
}

how to count up to a specific number in javascript, fizzbuzz

I'm trying to solve a problem similar to the popular fizzbuzz or pingpong question. I already figured out how to write this portion of the code, however I also need to make a counter in which the user inputs their number and the page gives back 1 to the number they input, as well as the fizzbuzz/popcorn numbers. I think the code is something along the lines of
for (i = 1; i <= 20; i++) {...}
But I don't know how to combine this with my other code
var pingpong = function(number) {
if (number % 15 === 0) {
return 'pingpong';
} else if (number % 5 === 0) {
return 'pong';
} else if (number % 3 === 0) {
return 'ping';
} else {
return false;
}
};
Sorry, I know this is a super beginner question, but I'm just starting out and am having a hard time figuring out how everything works together.

You are close. To read more about JavaScripts loops.
var pingpong = function(number) {
if (number % 15 === 0) {
return 'pingpong';
} else if (number % 5 === 0) {
return 'pong';
} else if (number % 3 === 0) {
return 'ping';
} else {
return false;
}
};
for (var i = 1; i <= 20; i++) {
var num = pingpong(i);
// Just show it on the screen
document.body.innerHTML += i + ': ' + num + '<br />';
}

FizzBuzz program (details given) in Javascript [closed]

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Can someone please correct this code of mine for FizzBuzz? There seems to be a small mistake. This code below prints all the numbers instead of printing only numbers that are not divisible by 3 or 5.
Write a program that prints the numbers from 1 to 100. But for multiples of three, print "Fizz" instead of the number, and for the multiples of five, print "Buzz". For numbers which are multiples of both three and five, print "FizzBuzz".
function isDivisible(numa, num) {
if (numa % num == 0) {
return true;
} else {
return false;
}
};
function by3(num) {
if (isDivisible(num, 3)) {
console.log("Fizz");
} else {
return false;
}
};
function by5(num) {
if (isDivisible(num, 5)) {
console.log("Buzz");
} else {
return false;
}
};
for (var a=1; a<=100; a++) {
if (by3(a)) {
by3(a);
if (by5(a)) {
by5(a);
console.log("\n");
} else {
console.log("\n");
}
} else if (by5(a)) {
by5(a);
console.log("\n");
} else {
console.log(a+"\n")
}
}

for (let i = 1; i <= 100; i++) {
let out = '';
if (i % 3 === 0) out += 'Fizz';
if (i % 5 === 0) out += 'Buzz';
console.log(out || i);
}

/*Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”*/
var str="",x,y,a;
for (a=1;a<=100;a++)
{
x = a%3 ==0;
y = a%5 ==0;
if(x)
{
str+="fizz"
}
if (y)
{
str+="buzz"
}
if (!(x||y))
{
str+=a;
}
str+="\n"
}
console.log(str);
Your functions return falsy values no matter what, but will print anyway. No need to make this overly complicated.
fiddle: http://jsfiddle.net/ben336/7c9KN/

Was fooling around with FizzBuzz and JavaScript as comparison to C#.
Here's my version, heavily influenced by more rigid languages:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
if (result.length ===0) result = i;
console.log(result);
}
}
I like the structure and ease of read.
Now, what Trevor Dixon cleverly did is relay on the false-y values of the language (false , null , undefined , '' (the empty string) , 0 and NaN (Not a Number)) to shorten the code.
Now, the if (result.length ===0) result = i; line is redundant and the code will look like:
function FizzBuzz(aTarget) {
for (var i = 1; i <= aTarget; i++) {
var result = "";
if (i%3 === 0) result += "Fizz";
if (i%5 === 0) result += "Buzz";
console.log(result || i);
}
}
Here we relay on the || operator to say : "if result is false, print the iteration value (i)". Cool trick, and I guess I need to play more with JavaScript in order to assimilate this logic.
You can see other examples (from GitHub) that will range from things like :
for (var i=1; i <= 20; i++)
{
if (i % 15 == 0)
console.log("FizzBuzz");
else if (i % 3 == 0)
console.log("Fizz");
else if (i % 5 == 0)
console.log("Buzz");
else
console.log(i);
}
No variables here, and just check for division by 15,3 & 5 (my above one only divides by 3 & 5, but has an extra variable, so I guess it's down to microbenchmarking for those who care, or style preferences).
To:
for(i=0;i<100;)console.log((++i%3?'':'Fizz')+(i%5?'':'Buzz')||i)
Which does it all in on line, relaying on the fact that 0 is a false value, so you can use that for the if-else shorthanded version (? :), in addition to the || trick we've seen before.
Here's a more readable version of the above, with some variables:
for (var i = 1; i <= 100; i++) {
var f = i % 3 == 0, b = i % 5 == 0;
console.log(f ? b ? "FizzBuzz" : "Fizz" : b ? "Buzz" : i);
}
All in all, you can do it in different ways, and I hope you picked up some nifty tips for use in JavaScript :)

.fizz and .buzz could be CSS classes, no? In which case:
var n = 0;
var b = document.querySelector("output");
window.setInterval(function () {
n++;
b.classList[n%3 ? "remove" : "add"]("fizz");
b.classList[n%5 ? "remove" : "add"]("buzz");
b.textContent = n;
}, 500);
output.fizz:after {
content: " fizz";
color:red;
}
output.buzz:after {
content: " buzz";
color:blue;
}
output.fizz.buzz:after {
content: " fizzbuzz";
color:magenta;
}
<output>0</output>

With ternary operator it is much simple:
for (var i = 0; i <= 100; i++) {
str = (i % 5 == 0 && i % 3 == 0) ? "FizzBuzz" : (i % 3 == 0 ? "Fizz" : (i % 5 == 0) ? "Buzz" : i);
console.log(str);
}

for(i = 1; i < 101; i++) {
if(i % 3 === 0) {
if(i % 5 === 0) {
console.log("FizzBuzz");
}
else {
console.log("Fizz");
}
}
else if(i % 5 === 0) {
console.log("Buzz");
}
else {
console.log(i)
}
}

In your by3 and by5 functions, you implicitly return undefined if it is applicable and false if it's not applicable, but your if statement is testing as if it returned true or false. Return true explicitly if it is applicable so your if statement picks it up.

As an ES6 generator: http://www.es6fiddle.net/i9lhnt2v/
function* FizzBuzz() {
let index = 0;
while (true) {
let value = ''; index++;
if (index % 3 === 0) value += 'Fizz';
if (index % 5 === 0) value += 'Buzz';
yield value || index;
}
}
let fb = FizzBuzz();
for (let index = 0; index < 100; index++) {
console.log(fb.next().value);
}

Codeacademy sprang a FizzBuzz on me tonight. I had a vague memory that it was "a thing" so I did this. Not the best way, perhaps, but different from the above:
var data = {
Fizz:3,
Buzz:5
};
for (var i=1;i<=100;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}
It relies on data rather than code. I think that if there is an advantage to this approach, it is that you can go FizzBuzzBing 3 5 7 or further without adding additional logic, provided that you assign the object elements in the order your rules specify. For example:
var data = {
Fizz:3,
Buzz:5,
Bing:7,
Boom:11,
Zing:13
};
for (var i=1;i<=1000;i++) {
var value = '';
for (var k in data) {
value += i%data[k]?'':k;
}
console.log(value?value:i);
}

This is what I wrote:
for (var num = 1; num<101; num = num + 1) {
if (num % 5 == 0 && num % 3 == 0) {
console.log("FizzBuzz");
}
else if (num % 5 == 0) {
console.log("Buzz");
}
else if (num % 3 == 0) {
console.log("Fizz");
}
else {
console.log(num);
}
}

for (var i = 1; i <= 100; i++) {
if (i % 3 === 0 && i % 5 === 0) console.log("FizzBuzz");
else if (i%3 === 0) console.log("Fizz");
else if (i%5 === 0) console.log("Buzz");
else console.log(i);
}
One of the easiest way to FizzBuzz.
Multiple of 3 and 5, at the same time, means multiple of 15.
Second version:
for (var i = 1; i <= 100; i++) {
if (i % 15 === 0) console.log("FizzBuzz");
else if (i%3 === 0) console.log("Fizz");
else if (i%5 === 0) console.log("Buzz");
else console.log(i);
}

In case someone is looking for other solutions: This one is a pure, recursive, and reusable function with optionally customizable parameter values:
const fizzBuzz = (from = 1, till = 100, ruleMap = {
3: "Fizz",
5: "Buzz",
}) => from > till || console.log(
Object.keys(ruleMap)
.filter(number => from % number === 0)
.map(number => ruleMap[number]).join("") || from
) || fizzBuzz(from + 1, till, ruleMap);
// Usage:
fizzBuzz(/*Default values*/);
The from > till is the anchor to break the recursion. Since it returns false until from is higher than till, it goes to the next statement (console.log):
Object.keys returns an array of object properties in the given ruleMap which are 3 and 5 by default in our case.
Then, it iterates through the numbers and returns only those which are divisible by the from (0 as rest).
Then, it iterates through the filtered numbers and outputs the saying according to the rule.
If, however, the filter method returned an empty array ([], no results found), it outputs just the current from value because the join method at the end finally returns just an empty string ("") which is a falsy value.
Since console.log always returns undefined, it goes to the next statement and calls itself again incrementing the from value by 1.

A Functional version of FizzBuzz
const dot = (a,b) => x => a(b(x));
const id = x => x;
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dot(f(3, 'fizz'), f(5, 'buzz')) (id) (n);
}
for more options in the above replace dot with dots as below
const dots = (...a) => f0 => a.reduceRight((acc, f) => f(acc), f0);
function fizzbuzz(n){
const f = (N, m) => n % N ? id : x => _ => m + x('');
return dots(f(3, 'fizz'), f(5, 'buzz'), f(7, 'bam')) (id) (n);
}
Reference: FizzBuzz in Haskell by Embedding a Domain-Specific Language
by Maciej Piro ́g

for (i=1; i<=100; i++) {
output = "";
if (i%5==0) output = "buzz";
if (i%3==0) output = "fizz" + output;
if (output=="") output = i;
console.log(output);
}

Functional style! JSBin Demo
// create a iterable array with a length of 100
// and map every value to a random number from 1 to a 100
var series = Array.apply(null, Array(100)).map(function() {
return Math.round(Math.random() * 100) + 1;
});
// define the fizzbuzz function which takes an interger as input
// it evaluates the case expressions similar to Haskell's guards
var fizzbuzz = function (item) {
switch (true) {
case item % 15 === 0:
console.log('fizzbuzz');
break;
case item % 3 === 0:
console.log('fizz');
break;
case item % 5 === 0:
console.log('buzz');
break;
default:
console.log(item);
break;
}
};
// map the series values to the fizzbuzz function
series.map(fizzbuzz);

Another solution, avoiding excess divisions and eliminating excess spaces between "Fizz" and "Buzz":
var num = 1;
var FIZZ = 3; // why not make this easily modded?
var BUZZ = 5; // ditto
var UPTO = 100; // ditto
// and easily extended to other effervescent sounds
while (num < UPTO)
{
var flag = false;
if (num % FIZZ == 0) { document.write ("Fizz"); flag = true; }
if (num % BUZZ == 0) { document.write ("Buzz"); flag = true; }
if (flag == false) { document.write (num); }
document.write ("<br>");
num += 1;
}
If you're using using jscript/jsc/.net, use Console.Write(). If you're using using Node.js, use process.stdout.write(). Unfortunately, console.log() appends newlines and ignores backspaces, so it's unusable for this purpose. You could also probably append to a string and print it. (I'm a complete n00b, but I think (ok, hope) I've been reasonably thorough.)
"Whaddya think, sirs?"

check this out!
function fizzBuzz(){
for(var i=1; i<=100; i++){
if(i % 3 ===0 && i % 5===0){
console.log(i+' fizzBuzz');
} else if(i % 3 ===0){
console.log(i+' fizz');
} else if(i % 5 ===0){
console.log(i+' buzz');
} else {
console.log(i);
}
}
}fizzBuzz();

Slightly different implementation.
You can put your own argument into the function. Can be non-sequential numbers like [0, 3, 10, 1, 4]. The default set is only from 1-15.
function fizzbuzz (set) {
var set = set ? set : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
var isValidSet = set.map((element) => {if (typeof element !== 'number') {return false} else return true}).indexOf(false) === -1 ? true : false
var gotFizz = (n) => {if (n % 3 === 0) {return true} else return false}
var gotBuzz = (n) => {if (n % 5 === 0) {return true} else return false}
if (!Array.isArray(set)) return new Error('First argument must an array with "Number" elements')
if (!isValidSet) return new Error('The elements of the first argument must all be "Numbers"')
set.forEach((n) => {
if (gotFizz(n) && gotBuzz(n)) return console.log('fizzbuzz')
if (gotFizz(n)) return console.log('fizz')
if (gotBuzz(n)) return console.log('buzz')
else return console.log(n)
})
}

var num = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var runLoop = function() {
for (var i = 1; i<=num.length; i++) {
if (i % 5 === 0 && i % 3 === 0) {
console.log("FizzBuzz");
}
else if (i % 5 === 0) {
console.log("Buzz");
}
else if (i % 3 === 0) {
console.log("Fizz");
}
else {
console.log(i);
}
}
};
runLoop();

Just want to share my way to solve this
for (i = 1; i <= 100; i++){
if (i % 3 === 0 && i % 5 === 0) {
console.log('fizzBuzz');
} else if (i % 3 === 0) {
console.log('fizz');
} else if (i % 5 === 0){
console.log('buzz');
} else {
console.log(i);
}
}

var limit = prompt("Enter the number limit");
var n = parseInt(limit);
var series = 0;
for(i=1;i<n;i++){
series = series+" " +check();
}
function check() {
var result;
if (i%3==0 && i%5==0) { // check whether the number is divisible by both 3 and 5
result = "fizzbuzz "; // if so, return fizzbuzz
return result;
}
else if (i%3==0) { // check whether the number is divisible by 3
result = "fizz "; // if so, return fizz
return result;
}
else if (i%5==0) { // check whether the number is divisible by 5
result = "buzz "; // if so, return buzz
return result;
}
else return i; // if all the above conditions fail, then return the number as it is
}
alert(series);

Thats How i did it :
Not the best code but that did the trick
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
for(var i = 0 ; i <= 19 ; i++){
var fizz = numbers[i] % 3 === 0;
var buzz = numbers[i] % 5 === 0;
var fizzBuzz = numbers[i] % 5 === 0 && numbers[i] % 3 === 0;
if(fizzBuzz){
console.log("FizzBuzz");
} else if(fizz){
console.log("Fizz");
} else if(buzz){
console.log("Buzz");
} else {
console.log(numbers[i]);
}
}

As much as this is easy logic it can be a daunting task for beginners. Below is my solution to the FizzBuzz problem:
let i = 1;
while(i<=100){
if(i % 3 ==0 && i % 5 == 0){
console.log('FizzBuzz');
}
else if(i % 3 == 0){
console.log('Fizz');
}
else if(i % 5 == 0){
console.log('Buzz');
}
else{
console.log(i);
}
i++;
}

considering performance and readability, please find my take on this problem
way 1: instead of doing a math modules operation in an if loop, which results in performing 3 times taking it a step above reduces the overhead
function fizzBuzz(n) {
let count =0;
let x = 0;
let y = 0;
while(n!==count)
{
count++;
x = count%3;
y = count%5;
if(x === 0 && y ===0)
{
console.log("fizzbuzz");
}
else if(x === 0)
{
console.log("fizz");
}
else if(y === 0)
{
console.log("buzz");
}
else
{
console.log(count);
}
}
}
fizzBuzz(15);
way 2: condensing the solution
function fizzBuzz(n) {
let x = 0;
let y = 0;
for (var i = 1; i <= n; i++) {
var result = "";
x = i%3;
y = i%5;
if (x === 0 && y === 0) result += "fizzbuzz";
else if (x === 0) result += "fizz";
else if (y === 0) result += "buzz";
console.log(result || i);
}
}
fizzBuzz(5)

Here's my favorite solution. Succinct, functional & fast.
const oneToOneHundred = Array.from({ length: 100 }, (_, i) => i + 1);
const fizzBuzz = (n) => {
if (n % 15 === 0) return 'FizzBuzz';
if (n % 3 === 0) return 'Fizz';
if (n % 5 === 0) return 'Buzz';
return n;
};
console.log(oneToOneHundred.map((i) => fizzBuzz(i)).join('\n'));

function fizzBuzz(n) {
for (let i = 1; i < n + 1; i++) {
if (i % 15 == 0) {
console.log("fizzbuzz");
} else if (i % 3 == 0) {
console.log("fizz");
} else if (i % 5 == 0) {
console.log("buzz");
} else {
console.log(i);
}
}
}
fizzBuzz(15);

Different functional style -- naive
fbRule = function(x,y,f,b,z){return function(z){return (z % (x*y) == 0 ? f+b: (z % x == 0 ? f : (z % y == 0 ? b: z))) }}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule(3,5, "fizz", "buzz"))
or, to incorporate structures as in above example: ie [[3, "fizz"],[5, "buzz"], ...]
fbRule = function(fbArr,z){
return function(z){
var ed = fbArr.reduce(function(sum, unit){return z%unit[0] === 0 ? sum.concat(unit[1]) : sum }, [] )
return ed.length>0 ? ed.join("") : z
}
}
range = function(n){return Array.apply(null, Array(n)).map(function (_, i) {return i+1;});}
range(100).map(fbRule([[3, "fizz"],[5, "buzz"]]))
OR, use ramda [from https://codereview.stackexchange.com/questions/108449/fizzbuzz-in-javascript-using-ramda ]
var divisibleBy = R.curry(R.compose(R.equals(0), R.flip(R.modulo)))
var fizzbuzz = R.map(R.cond([
[R.both(divisibleBy(3), divisibleBy(5)), R.always('FizzBuzz')],
[divisibleBy(3), R.aklways('Fizz')],
[divisibleBy(5), R.always('Buzz')],
[R.T, R.identity]
]));
console.log(fizzbuzz(R.range(1,101)))