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?

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


2 Answers

Alena Holligan
Alena Holligan
Treehouse Teacher

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

Henrik Christensen
Henrik Christensen
Python Web Development Techdegree Student 38,317 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:

echo $flavor;