Welcome to the Treehouse Community

The Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.

PHP Build a Simple PHP Application Creating the Menu and Footer Variables and Conditionals

Pascal Kellenberger
Pascal Kellenberger
20,410 Points

Answer to "Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears."

<?php
$flavor = "banana";
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
if ($flavor == "cookie dough.") {
echo "<p>Randy's favorite flavor is cookie dough, also!</p>";
}
?>

Here is my (wrong?) answer:

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

3 Answers

David Kaneshiro Jr.
David Kaneshiro Jr.
29,247 Points

Hi Pascal Kellenberger,

If you look closely at your "cookie dough" answer in your original post you will see that you have a '.' within the string literal value for $flavor. The extra '.' at the end of the string "cookie dough" is what caused the answer in your original post to fail. So the second answer didn't pass because you used single quotes instead of double quotes. It passed because the '.' was no longer present in the "cookie dough" string literal.

I hope this clears things up.

Pascal Kellenberger
Pascal Kellenberger
20,410 Points

This one worked:

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

A bug? I don't see a difference... except the '' instead of ""

Pascal Kellenberger
Pascal Kellenberger
20,410 Points

Thank you for your answer! I tried it again and you're right - it's the "."