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![Backston Makuve](https://secure.gravatar.com/avatar/bafbe1356423ade3c64d40b723f44f3c?s=96&d=https%3A%2F%2Fecs-static.teamtreehouse.com%2Fassets%2Fcontent%2Fdefault_avatar-445fbbabfc8dc9188fb5967fe43322ee0c3e0dd1e10f378bf8343784af5a13eb.webp&r=pg)
Backston Makuve
951 Pointsconditional around the final echo command that checks if the variable has a value of "cookie dough"
The message in the final echo command only makes sense if your favorite flavor is the same as Hal's.Add a conditional around that final echo command that checks if the flavor variable has a value of "cookie dough."(Remember to choose carefully between using a equal sign and a double equal sign in the check.Preview code and if u have different flavor, make sure the message disapears.
<?php
$flavor='vanilla';
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor='vanilla';
echo ".</p>";
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
echo .Hal.$flavor='cookie dough,also!';
if($flavor=='.Hal.$flavor,also!';){
]
?>
1 Answer
![Simon Coates](https://secure.gravatar.com/avatar/9c452f3f160ef69f85f1ff042db63980?s=96&d=https%3A%2F%2Fecs-static.teamtreehouse.com%2Fassets%2Fcontent%2Fdefault_avatar-445fbbabfc8dc9188fb5967fe43322ee0c3e0dd1e10f378bf8343784af5a13eb.webp&r=pg)
Simon Coates
28,694 Pointsthe correct code is
<?php
$flavor = "Strawberry";
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>";
}
?>