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JavaScript JavaScript Basics (Retired) Making Decisions with Conditional Statements Using Comparison Operators

Posted by Dominique Armstrong
Dominique Armstrong
3,559 Points

The else clause is not being recognized with the alert command

I followed the instructions and now the challenge is telling me that the else clause alert is not there, when I clearly put it there.

I even put the curly brackets around the else clause to see if that was the problem but it made things worse so I left them out. Is there something I'm missing to solve this problem. If there is, could you explain what I did wrong so I can understand?

Thank you kindly!

script.js 
var a = 10;
var b = 20;
var c = 30;

if ('a > b')
  alert ('a is greater than b'); 

else ('a < b')
 alert ('a is not greater than b');
index.html 
<!DOCTYPE HTML>
<html>
<head>
  <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
  <title>JavaScript Basics</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>

2 Answers

Steven Parker
Steven Parker
231,275 Points
Steven Parker
Steven Parker
231,275 Points

The message may be a bit misleading, the actual issue is that an "else" does not take a comparison expression. It doesn't need one since it handles anything not covered by the "if".

Dominique Armstrong
Dominique Armstrong
3,559 Points

Yeah that's what I thought but if I remove the comparison expression, I get this error message:

"I don't see the message 'a is NOT greater than b'. Are you sure you wrote the conditional statement correctly?"

Steven Parker
Steven Parker
231,275 Points
Steven Parker
Steven Parker
231,275 Points

I missed it the first time, but you don't want to put quotes around your conditional expression. That would make it a string literal, which by the fact that it's not empty is always "truthy".

Dominique Armstrong
Dominique Armstrong
3,559 Points
Dominique Armstrong
Dominique Armstrong
3,559 Points

Thank you Steven, that helped when I removed the quotes. I'm still having a hard time determining when to or not to put quotes in something. Thanks for your help!

Steven Parker
Steven Parker
231,275 Points
Steven Parker
Steven Parker
231,275 Points

This hint might help: Generally, you'll never put quotes around a variable name, or an expression that uses an operator for calculating or comparing anything. Putting quotes around something makes a string literal, which doesn't represent anything except for the actual letters and symbols in the quotes. For example:

console.log("23 + 17");  // the console will show:  23 + 17
console.log(23 + 17);    // the console will show:  40