Build a simple PHP application: task 4 of 4

Question

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

When i look in the editor the right message shows up but it wont let me pass this task, what am i missing?

Answer 1 · 2014-10-24T14:14:44Z

October 24, 2014 2:14pm

Hi Cassandra,

Looks like you left out some stuff from prior questions and you've read too much into the question. Take another look at the question then study the code below, and, what Mark said.

Jeff

<?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>";
}

?>

Answer 2 · 2014-10-24T14:08:04Z

October 24, 2014 2:08pm

Hi Cassandra!

You'll have to echo some kind of a string to display nothing in the else portion of your if conditional after changing the flavor variable to "cookie dough". Unfortunately, I can't tell you a whole lot more without giving the answer away completely. You're really close though!

Answer 3 · 2014-10-24T14:17:59Z

October 24, 2014 2:17pm

Ahh now i got it, thank you Mark and Jeff!!

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Build a simple PHP application: task 4 of 4

3 Answers

Jeff Busch

Jeff Busch

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