Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears.

hmm i see you have a space before the word in your $flavor variable. That could be it. try this:

<?php
$flavor = "cookie dough";
echo "<p>Your favorite flavor of ice cream is $flavor";
echo ".</p>";
if ($flavor == "cookie dough"){
echo "<p>Randy's favorite flavor is cookie dough, also!</p>";
}
?>

where is your if statement from task 3? please format your code in your question using markdown. also one error I spot is

$Flavor == "cookie dough"

you used $Flavor when you declared your variable earlier as $flavor. change it to a lower case f and see if it works

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

whats wrong now??

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

whats wrong now??

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

whats wrong now??

you are supposed the change the value of your flavor variable to cookie dough to test the if conditional.

so change

$flavor = "vanilla";

to

$flavor = "cookie dough";

every time i change it it says that Task 2 is no longer acceptable..

thank you!