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Luciano Cartaxo
1,231 PointsBummer! You need to 'echo' the value in $flavor.
Bummer! You need to 'echo' the value in $flavor.
<?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>";
?>
2 Answers
alastair cooper
30,617 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
21,074 PointsYou 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.
alastair cooper
30,617 Pointsalastair cooper
30,617 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>"; }
?>