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PHP Build a Simple PHP Application Creating the Menu and Footer Variables and Conditionals

Posted by Fatni Anass
Fatni Anass
Courses Plus Student 304 Points

Help

Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears. i dont understant this question ? i make that's code but they sey "" Oops! It looks like Task 3 is no longer passing.""

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

4 Answers

Adam Moore
Adam Moore
21,956 Points
Adam Moore
Adam Moore
21,956 Points

Your code:

<?php $flavor = "cookie dough";
echo $flavor;
if($flavor == "cookie dough" ){
echo "Your favorite flavor of ice cream is "; echo $flavor; echo ".";
 } ?>

May need to use concatenation with . instead of separate semi-colons for the statement to echo in the "if" statement. Like this:

<?php $flavor = "cookie dough";
echo $flavor;
if($flavor == "cookie dough" ){
echo "Your favorite flavor of ice cream is " .  $flavor . ".";
 } ?>
Fatni Anass
Fatni Anass
Courses Plus Student 304 Points

Challenge task 3 of 4 im doing great!

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

Challenge task 4 of 4 Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears.

Adam Moore
Adam Moore
21,956 Points
Adam Moore
Adam Moore
21,956 Points

Your variable looks like it has a "." in it, and I think it is just supposed to be $flavor = "cookie dough", not $flavor = "cookie dough."

Fatni Anass
PLUS
Fatni Anass
Courses Plus Student 304 Points
Fatni Anass
Fatni Anass
Courses Plus Student 304 Points

oh yes thank you so much for your answers i try but it not work !! this the code :

<?php $flavor = "cookie dough"; echo "<p>Your favorite flavor of ice cream is "; echo $flavor; echo ".</p>"; echo $flavor;

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

Adam Moore
Adam Moore
21,956 Points
Adam Moore
Adam Moore
21,956 Points

The challenge asks for you to put the last statement into a conditional, so that if the variable $flavor is equal to "cookie dough," then it will echo out "Randy's favorite flavor is cookie dough, also!", not simply echo out the variable $flavor. Like this: 

<?php

$flavor = "cookie dough";

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

if($flavor == "cookie dough"){
    echo "<p>Randy's favorite flavor is cookie dough, also!</p>";
}
?>
Fatni Anass
Fatni Anass
Courses Plus Student 304 Points

Thank you so mush Adam it's work :D