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JavaScript JavaScript Basics (Retired) Storing and Tracking Information with Variables Using String Methods

JavaScript code challenge

Complete the assignment to the userName variable by adding a # symbol followed by an all uppercase version of the lastName variable. In other words, using string concatenation so that the final value of userName is "23188XTR#SMITH".

I don't know what I did wrong?

app.js
var id = "23188xtr";
var lastName = "Smith";

var userName = id.toUpperCase() + "#" + lastname.toUpperCase();
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="app.js"></script>
</body>
</html>

3 Answers

Rich Donnellan
MOD
Rich Donnellan
Treehouse Moderator 27,696 Points

Edwin,

Check your variable names in all code — they are case sensitive!

-Rich

May I know what you mean by case sensitive?

Sam Baines
Sam Baines
4,315 Points

Rich means that the variable names must match completely including where they use upper case and lower case letters. Case sensitive means variable 'ID' is not equal to 'id' because one is in lower case and one in upper case.

TJ Egan
TJ Egan
14,420 Points

Should be:

var id = "23188xtr";
var lastName = "Smith";

var userName = id.toUpperCase() + "#" + lastName.toUpperCase();

You had last*name.toUpperCase(), should have been lastN*ame.toUpperCase

Rich Donnellan
MOD
Rich Donnellan
Treehouse Moderator 27,696 Points

Sure!

variableName is not the same as variablename.

Got it?