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PHP

Jorge Alvarez
Jorge Alvarez
9,330 Points
Posted by Jorge Alvarez
Jorge Alvarez
9,330 Points

code challenge

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

Now this code actually passed me to the next task except that next task contradicts the previous one so it doesn't let me pass "Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears." I got an error saying that now the previous task isn't working so is it my code or is the program confused?

1 Answer

Phil White
PLUS
Phil White
Courses Plus Student 9,519 Points
Phil White
Phil White
Courses Plus Student 9,519 Points

Got the answer to both:

Task 3 

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

Task 4 

<?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>";
}

?>

Make sure you indent the last echo as above.