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

Alec Jones
Alec Jones
5,510 Points
Posted by Alec Jones
Alec Jones
5,510 Points

My Code Doesnt Work

I cant seem to get this code to work. Can someone please show me what I did wrong!

<?php

$flavor="choco";

echo "Your favorite flavor of ice cream is ";

echo $flavor;

echo ".";

echo "Randy's favorite flavor is cookie dough, also!";

?>

2 Answers

Andrew Kiernan
Andrew Kiernan
26,892 Points
Andrew Kiernan
Andrew Kiernan
26,892 Points

Hi Alec!

I'm not sure which step of the challenge is tripping you up, but I assume its the step with the conditional? If that's the case, you want to wrap the final echo statement in an "if" statement. It will look like this:

if ($flavor == "cookie dough") {
  echo "Randy's favorite flavor is cookie dough, also!"
};
John Breiner
John Breiner
6,918 Points
John Breiner
John Breiner
6,918 Points

Andrew is on the right part for the conditional, I thought I'd show you a cleaner way to do the code for the other statement.

echo "Your favorite flavor of ice cream is " . $flavor . ".";