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Enhancing PhP Challenge

The flavor.php include file contains a function named get_flavor. That function receives no arguments, and it returns a piece of text as its return value. Call that function and assign the return value to a variable named flavor. (If someone could just fix this so I can move on that would be great. I have tried a lot of different things.)

<?php include("flavor.php"); function get_flavor(){

return $flavor; } getflavor(); echo "Randy's favorite flavor of ice cream is ____.";;


4 Answers

I figured it out:

<?php include("flavor.php");

$flavor = get_flavor();

echo "Randy's favorite flavor of ice cream is ____.";;


Caroline Hagan
Caroline Hagan
12,612 Points

I can see you have a double ;; in your code there...


function get_flavor(){

$flavor = "vanilla";

return $flavor;


echo "Randy's favorite flavor of ice cream is " . get_flavor();


Thanks for this. But I figured it out :)