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

Simon Strömberg
Simon Strömberg
1,423 Points

The message in the final echo command only makes sense if your favorite flavor is the same as mine. Add a conditional ar

http://puu.sh/eniwN/14cd229a09.png

I cant get this to work, help thank you!!!

index.php
<?php

$flavor = "vanilla"; 
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
}
?>

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

3 Answers

Chase Marchione
Chase Marchione
155,055 Points

Hi there Simon,

  • Looks like you have an extra '}' in there. (Line 6 in your code)
  • You want to change the initial variable declaration to "cookie dough". In other words, it should no longer be initialized as vanilla.
  • You have the 'your favorite flavor' statement in there twice.
  • Looks like there's a paragraph tag that needs to be closed.
  • One php tag can do it for ya.

Example code (you'll want to remove the extra spaces in the paragraph tags... I did that because the forum was treating them as if they were real tags):

<?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 >";

?>

Hope this helps!

Chase Marchione
Chase Marchione
155,055 Points

Be sure to remove the spaces in the p tags.

Edit: Glad it worked!

Simon Strömberg
Simon Strömberg
1,423 Points

Hello again, task 4 didnt work thought. How do i hide the cookie dough value if it's not correct?

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

Simon Strömberg
Simon Strömberg
1,423 Points

I now see that <p> and </p> was missing, thanks