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);
}
}
Related
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');
}
I've made a sequence of numbers from 0 to 20 and I want to change the sequence so once it comes up with a multiple of 3 and 5 it logs 'FizzBuzz' to the terminal then carries on with the rest of the numbers up to 20. My problem is once I have changed the number to the string the rest of the terms in the sequence come up with NaN. I know the problem with my code is that I'm changing the number to a string and you cannot perform addition to a string which is why NaN comes up. I'm pretty new to this so any thoughts on how to do complete this would be greatly appreciated. I've tried using .append() but I'm pretty sure I'm using it incorrectly.
My code thus far;
var increment = function(number)
{
for (var i = 1; i <= 20; i++)
{
console.log(number++)
if ((number % 3 === 0) && (number % 5 === 0))
{
number = "FizzBuzz"
console.log("FizzBuzz");
}
else if (number % 3 === 0)
{
console.log("Fizz");
}
else if (number % 5 === 0)
{
console.log("Buzz");
}
else
{}
}
}
increment(1)
When you find a multiple of 3 and 5, you are setting number to "FizzBuzz", which does not have a ++ operator. On the next iteration, you call ++ on number, which is now "FizzBuzz", so it logs NaN.
If you don't set number to "FizzBuzz" it should work fine.
This would do it
var increment = function()
{
for (var i = 1; i <= 20; 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)
}
}
}
increment()
There's no need to pass the number parameter, since your for loop is already set to increment by 1 (i++).
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++;
}
I am trying to iterate through the arrays in the numbers variable, and if a number can be divided by 3 I'm logging "fizz", if it can be divided by 5 I'm logging "buzz", and if a number can be divided by 3 + 5, or 15, I'm logging "fizzbuzz"
Here is the working code:
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 = 1; i <= numbers.length; i++) {
if (i % 15 === 0) {
console.log("FizzBuzz");
} else if (i % 5 === 0) {
console.log("Buzz");
} else if (i % 3 === 0) {
console.log("Fizz");
} else {
console.log(i);
}
};
Here is my original code, which doesn't log "fizzbuzz"
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 = 1; i <= numbers.length; i++) {
if (i % 5 === 0) {
console.log("Buzz");
} else if (i % 3 === 0) {
console.log("Fizz");
} else if (i % 15 === 0) {
console.log("FizzBuzz");
} else {
console.log(i);
}
};
Why does the (i % 15 === 0) condition need to precede the other two conditions? Shouldn't it not matter?
The number 15 is divisible by both 3 and 5. If you don't test it first, then you'll never get there.
So let's take 30 as an example. If you check 15 first, you'll see that it's divisible by 15. However, if you check either 5 or 3 first, it'll be flagged as being divisible by either of those.
After one of your conditions evaluates as true, you break out of that if-block and don't evaluate following else-if/else statements. If you want the rest of them to evaluate you can make them if statements instead of else-if's.
It's because firstly the computer checks for the first statement and then the others as you use the ELSE IF which means, to check the statement after the first one is false. Use IF for all instead.