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PHP

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

PHP Development challenge task 3 of 4

im stuck on this question:

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.

What am i doing wrong?

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

3 Answers

Stone Preston
Stone Preston
42,016 Points
Stone Preston
Stone Preston
42,016 Points

you have your flavor variable in quotes:

echo "$flavor";

remove the quotes so that its just echo $flavor. you also probably need to keep the <p> tags in the echo statements, you removed them. other than that your code looks good. you should have:

<?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!"; 
} 
?>
Phil White
Phil White
Courses Plus Student 9,519 Points
Phil White
Phil White
Courses Plus Student 9,519 Points

Thank you it worked! But stuck on task 4 now:

Change the value in the flavor variable to cookie dough. Preview the code and make sure the message appears.

Tried this: 

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

But get error

Stone Preston
Stone Preston
42,016 Points
Stone Preston
Stone Preston
42,016 Points

ok you need to change your echo statement to echo the variable, not vanilla. also indent your echo command inside the if statement. for some reason it wont work if its not indented in:

<?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>";
}

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

Done it! Thanks Stone and john the help is really appreciated!

John MacDonald
John MacDonald
8,508 Points
John MacDonald
John MacDonald
8,508 Points

Hey Phil

Make sure you echo out the variable $flavor where you are echoing out vanilla

Hope this helps

John; \n\n

James Gill
PLUS
James Gill
Courses Plus Student 34,935 Points
James Gill
James Gill
Courses Plus Student 34,935 Points

Yep--what Stone said. The devil is always in the details.