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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
Posted by Don Shipley
Don Shipley
19,488 Points

Modify the command that displays the flavor of ice cream. Instead of it displaying a static piece of text, change it to

<?php $flavor = "vanilla"; // if I change the variable here test 3 will no longer pass. echo "<p>Your favorite flavor of ice cream is "; echo $flavor; echo ".</p>";

// If I place a variable here to change to cookie dough task three will not pass. if($flavor == "cookie dough"){ echo "<p>Randy's favorite flavor is $flavor , also!</p>"; }

?> <p> <?php // if I add a new block and the variable of cookie dough here task 2 will not pass. $flavor = "cookie dough"; echo "<p>Randy's favorite flavor is $flavor , also!</p>"; ?>

1 Answer

Gareth Borcherds
Gareth Borcherds
9,372 Points
Gareth Borcherds
Gareth Borcherds
9,372 Points

Here's what it looks like at the end

<?php
$flavor = '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 cookie dough, also!</p>";
}
?>
Don Shipley
Don Shipley
19,488 Points
Don Shipley
Don Shipley
19,488 Points

Instead of trying with the same block I added this to the end of the block.. Guess in the test question you need to add a whole new block so the variable will not change the first block. with just the $flavor = "cookie dough"; if($flavor == "cookie dough"){ echo "<p>Randy's favorite flavor is $flavor, also!</p>"; } which did not pass. Thank you just added yours to the end and passed.