Basic JavaScript: Use Bracket Notation to Find the Last Character in a String

JavaScript
1

Tell us what’s happening:

this is my code. I am stumped. can you please help me get the answer.

Your code so far
// Example
var firstName = “Ada”;
var lastLetterOfFirstName = firstName[firstName.length - 1];

// Setup
var lastName = “Lovelace”;

// Only change code below this line.
var lastLetterOfLastName = lastName[lastName.length - 1];

console.log(lastLetterofLastName);


// Example
var firstName = "Ada";
var lastLetterOfFirstName = firstName[firstName.length - 1];

// Setup
var lastName = "Lovelace";

// Only change code below this line.
var lastLetterOfLastName = lastName[lastName.length - 1];

console.log(lastLetterofLastName);

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/80.0.3987.116 Safari/537.36.

Challenge: Use Bracket Notation to Find the Last Character in a String

Link to the challenge:
https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-last-character-in-a-string

2

ReferenceError: lastLetterofLastName is not defined

That’s odd, isn’t it? Seems like you have to create that variable… I would definitely take a look at the variables that you’ve made and check to see if they are spelled properly!

Best of luck!

3

Do I have to ass the string property? I am confused by the “defined” part.

4

use* I apologize

this is just some extra characters

5

When you define a variable you will specify the type and then the variable name.

var ImAVariable = 1;
const ImAlsoAVariable = "Hello";
let ITooAmAVariable = "I'm a variable!";
console.log(NonExistingVariable)

You can see that I’ve defined the variables ImAVariable, ImAlsoAVariable, and ITooAmAVariable. When I try to console.log() a variable that I haven’t created it lets us know that the variable hasn’t been defined. In our case here,

ReferenceError: lastLetterofLastName is not defined

You have no created the variable lastLetterofLastName. It’s important to remember that variable names are case sensitive.

var variable = 2;
var VARIABLE = 3;

These are considered two different variables.
I hope this brings you closer to a solution. Feel free to continue asking if you need further assistance.

6

You have a typo.
You declared the variable lastLetterOfLastName but then you attempted to log the variable lastLetterofLastName. Because these variable names are not exactly the same, you are getting an error that lastLetterofLastName is not defined.

7

Thank you all so much!!