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JavaScript JavaScript Loops, Arrays and Objects Simplify Repetitive Tasks with Loops Refactor Using a Loop

Chadrak Fertil
Chadrak Fertil
2,221 Points

Isn't this right?

Why wont this work?

script.js
console.log(2);
console.log(4);
console.log(6);
console.log(8);
console.log(10);
console.log(12);
console.log(14);
console.log(16);
console.log(18);
console.log(20);
console.log(22);
console.log(24);


forย (var i = 0; i <ย 25; i++2;) {
ย  console.log(i);
}
index.html
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>JavaScript Loops</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>

3 Answers

Darryl Mah
Darryl Mah
5,492 Points

The system probably doesnโ€™t like that the redundant console logs are still there. Delete them and your code should be good :)

Darryl Mah
Darryl Mah
5,492 Points

I didnโ€™t look at the code very well. Good catch Kris...also does ++2 work? Or does it need to be +=2? I thought ++ just meant increment by one?

Also your initial value to log is 2. And there shouldn't be a semicolon after i++2.

It should be +=2