6/6 What's wrong with my code?


"Oops, try again. It looks like you didn't log "Alexander Waumans" to the console. Did you remember to call list() after adding your new contact?" This is the error I receive.

function add(firstName, lastName, email, phoneNumber) {
    contacts[contacts.length] = {
    firstName: firstName,
    lastName: lastName,
    email: email,
    phoneNumber: phoneNumber  
add("Alexander", "Waumans", "alexander.waumans@example.com","123456789");


i ran your code, and the error i got is that your search function is missing (which you created in this exercise

if you need more help, post an updated version of your code


var bob = {
    firstName: "Bob",
    lastName: "Jones",
    phoneNumber: "(650) 777-7777",
    email: "bob.jones@example.com"

var mary = {
    firstName: "Mary",
    lastName: "Johnson",
    phoneNumber: "(650) 888-8888",
    email: "mary.johnson@example.com"

var contacts = [bob, mary];

function printPerson(person) {
    console.log(person.firstName + " " + person.lastName);

function list() {	
   var contactsLength = contacts.length; // i put var contactsLength here
   for (var i = 0; i < contactsLength; i++) {

/*Create a search function
then call it passing "Jones"*/

function add(firstName, lastName, phoneNumber, email) {
   contacts[contacts.length] = {
      firstName: firstName,
      lastName: lastName,
      phoneNumber: phoneNumber,
      email: email

add("Alexander", "Waumans", "123456789", "alexander.waumans@example.com");

As you can see, you problem was in scope of var contactsLength . In your case this variable was global and always returns 2.
After I replaced var contactsLength = contacts.length; inside function list() you will always have an actual contacts length.


Yes it works! Thank you!


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