Cant do 2/4

Modify the command that displays the flavor of ice cream. Instead of it displaying a static piece of text, change it to display the value stored in the 'flavor' variable.

<?php

$flavor ="";

echo "<p>Your favorite flavor of ice cream is ";

echo $flavor;

echo ".</p>";

echo "<p>Randy's favorite flavor is cookie dough, also!</p>";

?>

Since you are not using the markdown to display the coding properly, it is hard to read it laid out as it is. Here is what I used to pass it:

<?php
$flavor = vanilla;
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
echo "<p>Randy's favorite flavor is cookie dough, also!</p>";

?>

FYI, if you are interested, clicking the markdown cheatsheet (found below the reply box) can show you how to post the code properly.

Hello,

I think we overlooked the variable declaration in Peter's code. He is declaring the variable $flavor by assigning it a string. When you assign a string to anything in PHP you must surround it with quotes: either single quotes or double quotes. How Peter and Shane has their code, it would produce a syntax error. Add the quotes around vanilla and the code should pass.

Cheers!