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PHP

Posted by Simon DICKENS
Simon DICKENS
1,325 Points

PHP Code challange 3 of 4 Varibles and Condionals

The message in the final echo command only makes sense if your favorite flavor is the same as mine. 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 single equal sign and a double equal sign in the check.) Preview the code and, if you have a different flavor, make sure the message disappears.

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

cant get this right!

HELP PLEASE!

Simon

Drew S
Drew S
34,034 Points

Simon,

Delete the echo "on", and put echo "Randy's favorite flavor is cookie dough, also!"; inside the curly braces.

That should fix it for you.

Also, the echo "vanilla" you have should be echo $flavor.

2 Answers

Simon DICKENS
Simon DICKENS
1,325 Points
Simon DICKENS
Simon DICKENS
1,325 Points

Hi Andrew.

You mean like this?

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

Jacob Sørensen
Jacob Sørensen
3,768 Points
Jacob Sørensen
Jacob Sørensen
3,768 Points

This one worked for me:

<?php $flavor = "cookie dough";

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

?>