PHP Build a Basic PHP Website (2018) Building a Media Library in PHP Variables and Conditionals

Bummer! You need to 'echo' the value in $flavor.

Bummer! You need to 'echo' the value in $flavor.

index.php
<?php
$flavor = "vanilla";

echo ($flavor="vanilla");"<p>Your favorite flavor of ice cream is ";
echo "vanilla";
echo ".</p>";
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";

?>
alastair cooper
alastair cooper
28,196 Points

<?php

$flavor = "vanilla";

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

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

?>

echo vanilla using the variable, not a string! or concatenate it by doing... <?php $flavor = "vanilla";

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

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

?>

2 Answers

alastair cooper
alastair cooper
28,196 Points

$flavor = "vanilla";

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

if ($flavor=="cookie dough"); { echo "

Hal's favorite flavor is cookie dough, also!

"; } ?>

Carlos Federico Puebla Larregle
Carlos Federico Puebla Larregle
21,071 Points

You have to replace the string "vanilla" in the second string with your variable. In my case would be like this:

<?php
$flavor = "chocolate";

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

?>

I hope that helps a little bit.