Trying to execute FizzBuzz function.. why no work?

Question

function fizzBuzz (start, end) {
for ( var i = start; i <= end; 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);
        };
    };
};

fizzBuzz (1,10);

Trying to execute FizzBuzz function. I thought it was a syntax issue, maybe I'm overlooking something fundamental?

Your quotation marks that you're using for the strings aren't the proper character. You have `“` and it should be `"`. — Lye Fish, Aug 01 '15 at 04:33
You copy-pasted your code out of a source that allowed it's sample code to be broken. — Pointy, Aug 01 '15 at 04:33
possible duplicate of [Codecademy FizzBuzz app, stuck on step 1](http://stackoverflow.com/questions/8797834/codecademy-fizzbuzz-app-stuck-on-step-1) — Andrew Savinykh, Aug 01 '15 at 04:54

score 0 · Accepted Answer · answered Aug 01 '15 at 04:37

It is a syntax issue:

function fizzBuzz (start, end) {
    for ( var i = start; i <= end; 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);
        }
    }
}

fizzBuzz(1,10);

This is how it should look. You have unnecessary extra semicolons and wrong quotes, although only the latter truly broke your code.

Trying to execute FizzBuzz function.. why no work?

1 Answers1