javascript fizzbuzz switch statement

Question

I\'m currently taking the code academy course on Javascript and I\'m stuck on the FizzBuzz task. I need to count from 1-20 and if the number is divisible by 3 print fizz, by

Answer 1

Switch statement checks if the situation given in the cases matches the switch expression. What your code does is to compare whether x divided by 3 or 5 is equal to x which is always false and therefore the default is always executed. If you really want to use a switch statement here is one way you can do.

for (var i=1; i<=30; i++){
  switch(0){
    case (i % 15):
      console.log("fizzbuzz");
      break;
    case (i % 3):
      console.log("fizz");
      break;
    case (i % 5):
      console.log("buzz");
      break;
    default:
      console.log(i);
  }
}

Answer 2

I thought switch too,but no need.

    for (var n = 1; n <= 100; n++) {
  var output = "";  
  if (n % 3 == 0)
    output = "Fizz";
  if (n % 5 == 0)
    output += "Buzz";
  console.log(output || n);
}

Answer 3

The switch(true) part of this statement helped me. I was trying to do a switch statement for fizzbuzz. My solution incorporates the coding style of Rosettacodes - general solution. Most significantly the use of force typing to shorten the primary conditionals. I thought, it was valuable enough to post:

var fizzBuzzSwitch = function() {
  for (var i =0; i < 101; i++){
    switch(true) {
      case ( !(i % 3) && !(i % 5) ):
        console.log('FizzBuzz');
        break;
      case ( !(i % 3) ):
        console.log('Fizz');
        break;
      case ( !(i % 5) ):
       console.log('Buzz');
       break;
      default:
       console.log(i);
    }
  }
}

Answer 4

We can use a function to find a multiple of any number and declare two variables to identify these multiples so that if you want to change the multiples you only need to change at max 2 lines of code

function isMultiple(num, mod) {
    return num % mod === 0;
}
let a = 3;
let b = 5;
for(i=0;i<=100;i++){
    switch(true){
        case isMultiple(i,a) && isMultiple(i,b):
            console.log("FizzBuzz")
        case isMultiple(i,a):
            console.log("Fizz");
        case isMultiple(i,b):
            console.log("Buzz");
        default:
            console.log(i);
    }
}

Answer 5

Switch matches the x in switch(x){ to the result of evaluating the case expressions. since all your cases will result in true /false there is no match and hence default is executed always.

now using switch for your problem is not recommended because in case of too many expressions there may be multiple true outputs thus giving us unexpected results. But if you are hell bent on it :

for (var x = 0; x <= 20; x++) {
  switch (true) {
    case (x % 5 === 0 && x % 3 === 0):
        console.log("FizzBuzz");
        break;
    case x % 3 === 0:
        console.log("Fizz");
        break;
    case x % 5 === 0:
        console.log("Buzz");
        break;
    default:
        console.log(x);
        break;
  }

}

Answer 6

Here's a solution incorporating @CarLuvr88's answer and a switch on 0:

let fizzBuzz = function(min, max){
  for(let i = min; i <= max; i++){
    switch(0){
      case i % 15 : console.log('FizzBuzz'); break;
      case i % 3  : console.log('Fizz'); break;
      case i % 5  : console.log('Buzz'); break;
      default     : console.log(i); break;
    }
  }
}

fizzBuzz(1,20)