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Simingaye Dube
2,587 Pointsmodify the command ,change it to display the value stored in the 'flavor' variable.
Modify the command that displays the flavor of ice cream. Instead of it displaying a static piece of text, change it to display the value stored in the 'flavor' variable.
<?php
$flavor = "vanilla";
echo "<p>Your favorite flavor of ice cream is ";
echo "vanilla";
echo ".</p>";
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
?>
2 Answers
Cindy Lea
Courses Plus Student 6,497 PointsThey want you to store the value of $flavor in your message instead of using pre-defined value:
<?php $flavor = "vanilla"; echo "<p>Your favorite flavor of ice cream is "; echo $flavor; echo ".</p>"; echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
?>
Joy Manuel
6,464 Points<?php $flavor = "rocky road"; echo "<p>Your favorite flavor of ice cream is "; echo "vanilla"; echo ".</p>"; echo "<p>Hal's favorite flavor is cookie dough, also!</p>"; if ($flavor = "rocky road") { echo $flavor; }
?>
It says I passed, but this looks wrong to me.