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

Posted by Syed Mahbub
Syed Mahbub
8,780 Points

My code is rendering as it should be, but Im still getting an error.

So Task 3 is working fine,but in task 4, im changing the variable and for some reason that is causing "Task 3" to stop working.

Leonardo Hernandez
Leonardo Hernandez
13,798 Points 
<?php 

$flavor = "Cookie Dough";

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

 if ($flavor == "Cookie Dough") { 

echo "Randy's favorite flavor is cookie dough, also!"; 

} 

?>
Leonardo Hernandez
Leonardo Hernandez
13,798 Points 
<?php

 $flavor = "Cookie Dough"; 

echo "Your favorite flavor of ice cream is "; 

echo "$flavor";

echo "."; 

if ($flavor == "Cookie Dough") { 

echo "Randy's favorite flavor is cookie dough, also!";

 } 

?>

4 Answers

Louis Otto
Louis Otto
23,264 Points
Louis Otto
Louis Otto
23,264 Points

can you please post your code so we can take a look?

Syed Mahbub
Syed Mahbub
8,780 Points
Syed Mahbub
Syed Mahbub
8,780 Points

Hey Louis, here: <?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>"; } ?>

Louis Otto
Louis Otto
23,264 Points
Louis Otto
Louis Otto
23,264 Points

OK you've removed some of the <p> tags which are essential, but otherwise it's perfect. Remember also that it asks you to check what happens when it doesn't match, so change your flavor to chocolate as I've done, and the below will pass. I hope this makes sense:

<?php
$flavor = "chocolate";
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!"; 
    } ?>
?>
Leonardo Hernandez
Leonardo Hernandez
13,798 Points
Leonardo Hernandez
Leonardo Hernandez
13,798 Points

Sorry about the double post, the code I commented above is your code so that anyone trying to help can view it all easily.

I believe task one, two, and three concatenated some <p> tags into the echo commands. If you removed those, that could be your problem. In that case, your code as it functions is not the problem.

Louis Otto
Louis Otto
23,264 Points
Louis Otto
Louis Otto
23,264 Points

Also you wrapped the variable $flavor in quotes, you don't do that for variables, only strings.