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

Don Shipley
Don Shipley
19,488 Points

PHP test question

Where am I suppose to change the variable so vanilla and cookie dough work?

$flavor = "favorite"; echo "<p>Your favorite flavor of ice cream is "; echo $flavor = "vanilla"; echo ".</p>";

if ($flavor == "cookie dough") { echo "<p>Randy's favorite flavor is"; echo $flavor = "cookie dough" ; echo ", also!</p>"; }

2 Answers

Michal Kurtulík
Michal Kurtulík
9,203 Points
<?php
$flavor =  "ice cream"; // 4 setp. change to "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 $flavor, also!</p>";

?>
Don Shipley
Don Shipley
19,488 Points

Thank you for your help. Your code just makes the cookie dough line not appear. test code:

// echo "<p>Your favorite flavor of ice cream is "; // echo "vanilla"; // echo ".</p>"; // echo "<p>Randy's favorite flavor is cookie dough, also!</p>";

Here is what I have done with the test question. <?php first test question we make a variable named flavor $flavor ="whatever";

second question we need to change the text "vanilla" to read the $flavor variable answer echo $flavor = "vanilla";

third question Add a conditional around the final echo to check if the variable has a value of cookie dough.Preview the code and, if you have a different flavor, make sure the message disappears.

answer if($flavor == "cookie dough" )echo "<p>Randy's favorite flavor is $flavor, also!</p>";

Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears my answer will preview but not pass the test question. I think it is because of changing the global variable answer: $flavor = "cookie dough"; if ($flavor == "cookie dough"){ echo "<p>Randy's favorite flavor is"; echo $flavor; echo ", also!</p>"; ?>