PHP Build a Basic PHP Website (2018) Building a Media Library in PHP Variables and Conditionals

Tom Goldie
PRO
Tom Goldie
Pro Student 8,379 Points

Did what it asked, in ONE, and BOTH places, in all permutations. Checker does not accept answer.

This question is vague, for one, saying to replace the static that displays flavor with $flavor - and the "also" implies it should be replaced in both places. But I tried both, and even reformatted the first sentence onto a single line.

index.php
<?php

$flavor = "vanilla";
echo "<p>Your favorite flavor of ice cream is ";
echo "$flavor";
echo ".</p>";
echo "<p>Hal's favorite flavor is $flavor, also!</p>";

?>

2 Answers

You need to not wrap your variable name in a string:

<?php
$flavor = "vanilla";

echo "<p>Your favorite flavor of ice cream is ";
echo "$flavor"; // This is the line to change
echo ".</p>";
echo "<p>Hal's favorite flavor is $flavor, also!</p>"; // You also do not need to change this line
// Change the line above back to  "<p>Hal's favorite flavor is cookie dough, also!</p>"
?>

Try:

echo $flavor;

Not wrapping the variable name in the quotes allows the compiler to read it as a variable and pull out it's value

Tom Goldie
PRO
Tom Goldie
Pro Student 8,379 Points

Thanks. I discovered that on a fourth attempt, but asked the question in frustration after the third. I just tested it both ways on my WAMP server installation and they both work. Just a LITTLE frustrating that the answer parsing code doesn't accept both.

I appreciate you taking the time to help!