Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

PHP Build a Basic PHP Website (2018) Building a Media Library in PHP Variables and Conditionals

Stuart McDonald
Stuart McDonald
1,474 Points

Bug in echo flavor?

According to my understand, echo "$flavor"; should display the value of $flavor. The only way I can get this to work is echo $flavor; Shouldn't both be correct as they can both display the value of $flavor?

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

?>

2 Answers

Alena Holligan
STAFF
Alena Holligan
Treehouse Teacher

Stuart McDonald, you are correct either way should work and I have updated the code challenge :)

Henrik Christensen
seal-mask
.a{fill-rule:evenodd;}techdegree
Henrik Christensen
Python Web Development Techdegree Student 38,322 Points

You should get chocolate when you echo "$flavor"; since it's in double-quotes but there is no point doing it like that in this challenge.

Instead just do it like this:

<?php
echo $flavor;