
Garry Derby
342 Pointsvariables
$flavor = 'lemon sorbet';
<? php if ($flavor == "cookie dough") { echo "<p>Hal's favorite flavor is not cookie dough!</p>"; }
?>
<?php
$flavor = 'cookie dough';
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
echo "<p>Hal's favorite flavor is cookie dough! also!</p>";
?>
1 Answer

Cody Hansen
5,509 PointsI'm not entirely sure what you are asking here, but I have looked at it a while and come to the conclusion that the beginning code is your attempt for task #3 of the code challenge. If I am incorrect, just let me know and I'll do what I can to help.
Now, going off that assumption, let's look at your code! Currently, you have:
$flavor = 'lemon sorbet';
<? php if ($flavor == "cookie dough") { echo "<p>Hal's favorite flavor is not cookie dough!</p>"; }
?>
You're on the right track, Garry! Just a few minor fixes. Since the variable $flavor is a PHP variable, it must be inside the PHP tags like so:
<?php
$flavor = 'lemon sorbet';
if ($flavor == "cookie dough")
{
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
}
?>
The remaining code can also be placed between the same PHP tags to make it easier on you!
<?php
$flavor = 'lemon sorbet';
echo "<p>Your favorite flavor of ice cream is ";
echo $flavor;
echo ".</p>";
if ($flavor == "cookie dough")
{
echo "<p>Hal's favorite flavor is cookie dough, also!</p>";
}
?>
I hope this helps!
Garry Derby
342 PointsGarry Derby
342 PointsThanks a lot Cody. I had a brain freeze. I could see the answer but to type this out, was not happening.
Cody Hansen
5,509 PointsCody Hansen
5,509 PointsI'm glad to help! It happens to all of us.
Happy coding!