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PHP Enhancing a Simple PHP Application Integrating Validation Errors Reviewing PHP Basics

Ilya Liverts
Ilya Liverts
9,677 Points
Posted by Ilya Liverts
Ilya Liverts
9,677 Points

Can't solve second task.

<?php include("flavor.php"); var flavor = get_flavor(); echo "Randy's favorite flavor of ice cream is ____.";

?>

This code does not work. I can't find why.

2 Answers

thomascawthorn
thomascawthorn
22,986 Points
thomascawthorn
thomascawthorn
22,986 Points

To define you're variable, you'll want to use a dollar sign.

Spoiler

<?php
include('flavor.php');
$flavor = get_flavor();
echo "Randy's favorite flavor of ice cream is ____.";

?>
Fredrik August Madsen-Malmo
Fredrik August Madsen-Malmo
16,261 Points
Fredrik August Madsen-Malmo
Fredrik August Madsen-Malmo
16,261 Points

As said above, you can't declare variables in PHP using <code>var variable_name = value;</code>, this is JavaScript :) Instead you would use the <code>$variable_name = value</code> format.

Correct code

<?php
    include('flavor.php');
    $flavor = get_flavor();
    echo "Randy's favorite flavor of ice cream is ____.";
?>