使用if / else语句的函数

Question

I have created a function that will print all numbers between 1-24. If the number is dividable by 3, it will print Fizz, if dividable by 5, it will print Buzz.

If start > stop, I want to add an error message. How can I do that? I tried with alert but that did not work out at all.

2) When I print the function with start= 2 and stop=24 the first value that shows up is "undefined". Why is that?

By using .substr(1) I want to remove the first comma to the left.

function fizzBuzz(start, stop) {
  if (start >= stop)
 var returnString;
 for(var i= start; i <= stop; i++) {
    returnString += ",";
    if (i % 3 === 0 && i % 5 === 0)
      returnString += "FizzBuzz";
    else if (i % 3 === 0)
        returnString += "Fizz";
   else if (i % 5 === 0)
        returnString += "Buzz";
    else
        returnString += i;
}
 return returnString;
}
var result = fizzBuzz(2, 24).substr(1);

Thanks in advance!

Answer 1

You need to put validation in if clause

if (start >= stop)
{
   return "start must be less than stop";
}

And you need to initialize returnString with empty string first

var returnString = "";

Answer 2

You want to show error message if start >= stop , it can be done using alert . You are getting undefined for the first time because returnString variable didn't declare anywhere.

So try something like this. It would solve your problem

function fizzBuzz(start, stop) {
    var returnString = '';
    if (start >= stop) {
        alert('stop must be greater than start');
        return;
    }
    for (var i = start; i <= stop; i++) {
        returnString += ",";
        if (i % 3 === 0 && i % 5 === 0) 
            returnString += "FizzBuzz";
        else if (i % 3 === 0) 
            returnString += "Fizz";
        else if (i % 5 === 0) 
            returnString += "Buzz";
        else returnString += i;
    }
    return returnString;
}

JS Fiddle Demo

Answer 3

you can do this .

function fizzBuzz(start, stop) {
  if (start >= stop){
     throw "start must be less than stop";
  }
 var returnString = "";
  ...
  ...
  ...
 return returnString;
}
var result = fizzBuzz(2, 24).substr(1);

Answer 4

For the error mressage use throw

function fizzBuzz(start, stop) {
  if (start >= stop){
     throw "start must be less than stop";
  }
  ....

This requires you catch the error if you want to display info to the user

try{
    var result = fizzBuzz(72, 24).substr(1);
    alert(result);
}catch(e){
    alert(e); // alerts "start must be less than stop"
}

The second question is because resultString starts life as undefined and when you concatenate a string to undefined you get the string "UndefinedYOURSTRINGHERE". Define returnString as an empty string.

 var returnString = "";

Answer 5

The first value shows up is undefined because that's how you declared it:

if (start >= stop)
 var returnString;

Notice you don't have an opening bracket, so you basically just declare that variable. Even if you don't declare it, it will still be undefined later on when you do:

returnString += ",";

Since it always starts as undefined .

You can move that line to the end, and also you can add an alert at the first if check and return if start > stop :

function fizzBuzz(start, stop) {
    if (start >= stop) {
        alert('This is wrong');
        return;
    }


  returnString = "";
  for(var i= start; i <= stop; i++) {
    if (i % 3 === 0 && i % 5 === 0)
      returnString += "FizzBuzz";
    else if (i % 3 === 0)
        returnString += "Fizz";
   else if (i % 5 === 0)
        returnString += "Buzz";
    else
        returnString += i;

    returnString += ",";
}

 return returnString;
}
var res = fizzBuzz(2, 24);
var result = res.substring(0, res.length - 1);;
console.log(result);

Fiddle

Answer 6

You can use a throw statement for the error:

if (start >= stop){
    throw "first integer must be less than 2nd integer"
}

You must also declare returnString with an initial value:

 var returnString = "";

Answer 7

The problem is here,

if (start >= stop) 
var returnString;

There is no need for if statement here but still if you need you can do this,

  function fizzBuzz(start, stop) { if (start >= stop){
     var returnString;
 for(var i= start; i <= stop; i++) { returnString += ","; 
if (i % 3 === 0 && i % 5 === 0) returnString += "FizzBuzz"; 
else if (i % 3 === 0) returnString += "Fizz"; else if (i % 5 === 0) returnString += "Buzz";
 else returnString += i; }
 return returnString;
 }
    }
     var result = fizzBuzz(2, 24).substr(1);