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Pascal Kellenberger
20,410 PointsAnswer to "Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears."
<?php
$flavor = "banana";
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>";
}
?>
Here is my (wrong?) answer:
<?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>";
}
?>
3 Answers
David Kaneshiro Jr.
29,247 PointsIf you look closely at your "cookie dough" answer in your original post you will see that you have a '.' within the string literal value for $flavor. The extra '.' at the end of the string "cookie dough" is what caused the answer in your original post to fail. So the second answer didn't pass because you used single quotes instead of double quotes. It passed because the '.' was no longer present in the "cookie dough" string literal.
I hope this clears things up.
Pascal Kellenberger
20,410 PointsThis one worked:
<?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>";
}
?>
A bug? I don't see a difference... except the '' instead of ""
Pascal Kellenberger
20,410 PointsThank you for your answer! I tried it again and you're right - it's the "."