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Good afternoon,
The code below should represent the FizzBuzz game.
for (var i = 0, i < 100, i++) {
if(((i % 3) == 0) && ((i % 5) = 0)) {document.write('FizzBuzz')}
else if( ((i % 3) == 0) && ((i % 5) != 0)) {document.write('Fizz')}
else if( (( i % 3 ) != 0) && ((i % 5) == 0) ) {document.write('Buzz')}
else {document.write(i)}
}
This is the error I got in Mozilla Firefox Debugger
SyntaxError: missing ; after for-loop initialize 1.18.
I'm stuck.
There are two possible mistakes
1) Inside for loop conditional statement it should be ; but ,
2) There is a invalid left side assignment at ((i % 5) = 0), it should be ((i % 5) == 0)
To debug such issue use any linter and properly indent the code
for (var i = 0; i < 100; i++) {
if (((i % 3) == 0) && ((i % 5) == 0)) {
document.write('FizzBuzz')
} else if (((i % 3) == 0) && ((i % 5) != 0)) {
document.write('Fizz')
} else if (((i % 3) != 0) && ((i % 5) == 0)) {
document.write('Buzz')
} else {
document.write(i)
}
}
DEMO
You should use semi colons instead of commas in your loop:
for (var i = 0; i < 100; i++)
{
if ( i % 3 == 0 && i % 5 == 0 ) {document.write('FizzBuzz')}
else if ( i % 3 == 0 && i % 5 != 0 ) {document.write('Fizz')}
else if ( i % 3 != 0 && i % 5 == 0 ) {document.write('Buzz')}
else {document.write(i)}
}
You also have a massive parentheses overload! You don't need that many parens, they were also causing you errors.
Related
I can't get theright result, "Weird" on stdin, 18 and 20. Everything looks good to me, however something must be off.
if (N % 2 == 1) {
console.log("Weird");
}
else if ((N % 2 == 0) && (2 >= N <= 5)) {
console.log("Not Weird");
}
else if ((N % 2 == 0) && (5 <= N <= 20)) {
console.log("Weird");
}
else if ((N % 2 == 0) && (N > 20)) {
console.log("Not Weird");
}
else {
console.log("Weird");
}
'use strict';
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', inputStdin => {
inputString += inputStdin;
});
process.stdin.on('end', _ => {
inputString = inputString.replace(/\s*$/, '')
.split('\n')
.map(str => str.replace(/\s*$/, ''));
main();
});
function readLine() {
return inputString[currentLine++];
}
function main() {
const N = parseInt(readLine(), 10);
if (N%2==1) {
console.log("Weird");
}
else if ((N % 2 == 0) && (2 >= N <= 5)) {
console.log("Not Weird");
}
else if ((N % 2 == 0) && (5 <= N && N <= 20)) {
console.log("Weird");
}
else if ((N % 2 == 0) && (N > 20)) {
console.log("Not Weird");
}
else{
console.log("Weird");
}
}
I ve added the whole code. In the main function, in the second else if condition, there seems to be the problem. When n is given 18 or 20, I can not get the right output which should be "Weird"
You can't be doing two conditions in the same time
if (5<=N<=20) {}
Will evaluate 5<=N first which produces either true/false which are when compared to numbers will evaluate to (1/0) respectively. Then the second part ( <= 20) will be evaluated.
Combine two conditions only with AND / OR operators
if (5 <= N && N <= 20) {}
This will solve your problem.
I'm really novice to all of this, and I'm learning it in class. My assignment is to write an Else/If Loop where we display 1-100. If the integer is divisible by 3 display "play", if divisible by 4 display "ball", and if divisible by 3 and 4 display "Play Ball", anything else is just the integer.
I have my code working here, but I can't get my 3 and 4 to display "Play Ball", unless I run it by itself. Hope that makes sense, here's what I have:
for (var i = 1; i <= 100; i++) {
if (i % 3 === 0) {
console.log('Play');
} else if (i % 4 === 0) {
console.log('Ball');
} else if (i % 3 === 0 && i % 4 === 0) {
console.log('Play Ball');
} else {
console.log(i);
}
}
You just need to move thisi % 3 === 0 && i % 4 === 0 condition to the top:
for (var i = 1; i <= 100; i++) {
if (i % 3 === 0 && i % 4 === 0) {
console.log('Play Ball');
} else if (i % 3 === 0) {
console.log('Play');
} else if (i % 4 === 0) {
console.log('Ball');
} else {
console.log(i);
}
}
Since the first if statement will always be true when the third statement (with the &&) will be true, you'll never make it to the third statement. Try reordering as such:
for (var i = 1; i <= 100; i++) {
if (i % 3 === 0 && i % 4 === 0) {
console.log('Play Ball');
} else if (i % 4 === 0) {
console.log('Ball');
} else if (i % 3 === 0) {
console.log('Play');
} else {
console.log(i);
}
}
You can move the check for divisible by both 3 and 4 to the top:
for (var i = 1; i <= 100; i++) {
if (i % 3 === 0 && i % 4 === 0) {
console.log('Play Ball');
} else if (i % 3 === 0) {
console.log('Play');
} else if (i % 4 === 0) {
console.log('Ball');
} else {
console.log(i);
}
}
3 and 4 display
Above statement means, when BOTH condition satisfies then only you need to perform some task. In this scenario you need to use &&
if ( (i % 3 === 0) && (i % 4 === 0) ) {
console.log('Play Ball');
}
When to use else and/or else if
Let's take an example, you have a number, which is not divisible by 3 and 4 and you are not concern about that number. Then, you simply use else to fall all those category into it.
Now, there is another scenario, when you have entered number, which is again not divisible by 3 and 4. But, you might want see, if it is only divisible by 3 or 4. Then, you use else if where you can put the condition to check.
if ( (i % 3 === 0) && (i % 4 === 0) ) {
console.log('Play Ball');
}
else if ( i % 3 === 0) {
console.log('Play');
}
else if ( i % 4 === 0) {
console.log('Ball');
}
Hi I'm pretty new to coding. This fizzbuzz code isn't working any help would be greatly appreciated. I'm using sublime text 3 and trying to produce results on vagrant using the node command. Thank You - Saad
for(i=101; i<201; i++) {
if (i % 3 === 0 && i % 5 === 0){
console.log("FizzBuzz");
};
if(i % 3 === 0) {
console.log("Fizz");
} else if(i % 5 === 0) {
console.log("Buzz")
} else {
console.log(i);
}``
}
This could be done like this:
for(i = 101; i < 201; i++) {
var r = ""
if (i % 3 === 0){
r += "Fizz";
}
if(i % 5 === 0) {
r += "Buzz";
}
if(i % (5 * 3) !== 0){ // or just 15
r = i;
}
console.log(r);
}
please I'm stuck in this question below since yesterday. Below is the question:
Write a program that uses console.log to print all the numbers from 1
to 100, with two exceptions. For numbers divisible by 3, print "Fizz"
instead of the number, and for numbers divisible by 5 (and not 3), print
"Buzz" instead.
When you have that working, modify your program to print "FizzBuzz",
for numbers that are divisible by both 3 and 5 (and still print "Fizz" or
"Buzz" for numbers divisible by only one of those).
I only got the first two conditions but not the the third. I don't know how to go about it anymore, I've tried many options. Below is my code:
<html>
<head/head>
<body>
<script type="text/javascript">
for (i = 1; i <= 100; i++)
if (i % 3 == 0) {
document.write("Fizz");
document.write("<br />");
} else if (i % 5 == 0 && i % 3 != 0) {
document.write("Buzz");
document.write("<br />");
} else if (i % 3 && 5 == 0 && i % 3 != 0 && i % 5 != 0) {
document.write("FizzBuzz");
document.write("<br />");
} else {
document.write(+i);
document.write("<br />");
}
</script>
</body>
</html>
Check the most specific (FizzBuzz) condition first.
function fizzBuzz() {
for(var i = 1; 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);
}
}
}
here is an updated version of your code, I keep it as you write it with some changes, I made it work without touch it's logic, you can see that the problem was in the first comparison and in the second "if else" (5 will never be equal to 0). you can optimize the code more than that, good luck.
<html>
<head/head>
<body>
<script type="text/javascript">
for (i = 1; i <= 100; i++)
if (i % 3 == 0 && i % 5 != 0) {
document.write("Fizz");
document.write("<br />");
} else if (i % 5 == 0 && i % 3 != 0) {
document.write("Buzz");
document.write("<br />");
} else if (i % 3 == 0 && i % 5 == 0) {
document.write("FizzBuzz");
document.write("<br />");
} else {
document.write(+i);
document.write("<br />");
}
</script>
</body>
</html>
Since everyone is contributing, I might as well give you an interesting solution:
var i = 101;
while(i --> 0){ // as i goes to 0... wat
var state = !!(i % 3) << 1 | !!(i % 5), // compute state?
output = ["FizzBuzz", "Fizz", "Buzz", i]; // hmm...
console.log(output[state]); // output correct string
}
1st - Instead of document.write use console.log like the question says
2nd - You have a syntax error in the head section. it should be <head></head>
3rd - for the 1st part of the question all you need is this:
for (i = 1; i <= 100; i++) {
// if i is divisible by 3
if (i % 3 == 0) {
console.log("Fizz");
}
// if i is divisible by 5 (no need to check for 3 again)
else if (i % 5 == 0) {
console.log("Buzz");
}
// else
else {
console.log(i);
}
}
4th - For the 2nd part you need to add an extra if on top of what you have already:
for (i = 1; i <= 100; i++) {
// if i is divisible by 3 and 5
if (i % 3 == 0 && i % 5 == 0) {
console.log("FizzBuzz");
}
// if i is divisible by 3
else if (i % 3 == 0) {
console.log("Fizz");
}
// if i is divisible by 5 (no need to check for 3 again)
else if (i % 5 == 0) {
console.log("Buzz");
}
// else
else {
console.log(i);
}
}
working fiddle: https://jsfiddle.net/tedmanowar/amapqcLL/
You can use a while loop, than use a nested if statements to check the conditions.
let number = 0;
while (number <= 100) {
if(number % 3 === 0 && number % 5 === 0){
console.log("FizzBuzz");
}else if(number % 3 === 0){
console.log("Fizz");
}else if(number % 5 === 0){
console.log("Buzz");
}else{
console.log(number);
}
number++;
}
This solution is the easiest and simplest. There are multiple ways to solve the question though.
for (n=1; n<=100; n++){
let output = "";
if(n % 3=== 0) output += "Fizz"
if(n % 5=== 0) output += "buzz"
console.log(output || n);
}
I don't recommend this answer - since it is very hard to maintain - but it does do it in very few lines. It also relies on the two numbers only having a common factor of 1.
for(let i=1; i<=100; i++) {
console.log(`${i%15?i%5?i%3?i:'Fizz':'Buzz':'FizzBuzz'}`)
}
This is using the ternary operator and backquote template strings.
You should just use a loop that starts at 1 and is less than 101 (so up to 100) and test for the %n === 0. In other words, make sure there is no remainder.
function startConsoleDemo(){
for(var i=1,r; i<101; i++){ // loop from 1 to 100
r = i; // default value of r
if(i % 3 === 0 && i % 5 === 0){ // if i/3 and 1/5 do not produce a remainder
r = 'FizzBuzz'; // reassign r
}
else if(i % 3 === 0){ // we knew i % 5 !== 0 so see if i/3 does not produce a remainder
r = 'Fizz'; // reassign r
}
else if(i % 5 === 0){ // we already knew i % 3 !== 0 - you know the drill
r = 'Buzz'; // reassign r
}
console.log(r); // console at each step of the loop no matter what
}
}
startConsoleDemo(); // without () you can use like a var then () later
I have a solution and I'm pretty sure it'll work for you.
for (let i = 1; i <= 100; i++) {
if (i % 3 === 0 && i % 5 === 0) {
document.write(`${i} FizzBuzz`);
} else if (i % 3 === 0 && i % 5 !== 0) {
document.write(`${i} Fizz`);
} else if (i % 5 === 0 && i % 3 !== 0) {
document.write(`${i} Buzz`);
} else {
document.write(`${i}`);
}
}
This question already has answers here:
FizzBuzz program (details given) in Javascript [closed]
(27 answers)
Closed 4 years ago.
Here are Codecademy's instructions:
Print out the numbers from 1 - 20.
For numbers divisible by 3, print out "Fizz".
For numbers divisible by 5, print out "Buzz".
For numbers divisible by both 3 and 5, print out "FizzBuzz" in the
console.
Otherwise, just print out the number.
And here is my code:
for (i = 1; i <= 20; i++) {
if (i % 3 == 0) {
console.log("Fizz");
}
else if (i % 5 == 0) {
console.log("Buzz");
}
else if (i % 3 == 0 && i % 5 == 0) {
console.log("FizzBuzz");
}
else {
console.log(i);
}
}
The problem is that it won't print "FizzBuzz" for the number 15. It just prints "Fizz".
What am I missing here?
The else if only runs if all of the other statements so far have been false. Since i % 3 is true, the remaing else/if statements will never run. Try something like this:
for (i = 1; i <= 20; i++) {
if (i % 3 == 0 && i % 5 == 0) {
console.log("FizzBuzz");
}
else if (i % 5 == 0) {
console.log("Buzz");
}
else if (i % 3 == 0) {
console.log("Fizz");
}
else {
console.log(i);
}
}
If you want to use a nested conditional this seems to work.
for ( i=1 ; i < 21 ; i++){
if(i % 3 === 0) {
if(i % 5 === 0){
console.log("FizzBuzz");
}
else {
console.log("Fizz");
}
}
else if (i % 5 === 0) {
if ( i % 3 === 0){
console.log("FizzBuzz");
}
else{
console.log("Buzz");
}
}
else{
console.log(i);
}
}
straight forward fizzBuzz in javascript using tenary operators(a one-line shorthand for an if-else statement)
to learn about tenary operators go here
var i = 1;
while (i <= 20){
console.log((i % 3 === 0 && i % 5 === 0) ? "FizzBuzz" : (i % 3 === 0) ? "Fizz" : (i % 5 === 0 ? "Buzz" : i));
i++;
}