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Chris Hummel
13,960 Pointscode is right, but get error
I tested my code outside of this test, and it displays the flavors to the screen, but I still get an error saying I didn't use the same variable. What gives?
<?php
$flavors = array();
$flavors[] = array("name" => "Cookie Dough", "in_stock" => true);
$flavors[] = array("name" => "Vanilla", "in_stock" => false);
$flavors[] = array("name" => "Avocado Chocolate", "in_stock" => false);
$flavors[] = array("name" => "Bacon Me Crazy", "in_stock" => true);
$flavors[] = array("name" => "Strawberry", "in_stock" => false);
//add your code below this line
foreach ($flavors as $key => $value) {
echo $flavors[$key]["name"] . "<br />\n";
}
?>
1 Answer

Jordan Howlett
3,369 PointsIt seems like you may have overthought the solution, take a look at this piece of code and see if it helps clarify what they were asking for.
foreach ($flavors as $flavor) {
if($flavor["in_stock"] == true){
echo $flavor['name'];
}
}