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PHP Enhancing a Simple PHP Application Refactoring the Codebase Separating Concerns: MVC

Posted by Yasemin Sipahioglu
Yasemin Sipahioglu
Courses Plus Student 397 Points

refactoring the Codebase challenge 4/4

"<?php

// INCLUDE MODEL FILE require_once("flavors.php");

// CONTROLLER CODE $flavors = get_flavors_by_likes(3);

// VIEW CODE ?><html> <body>

<h1>Ye Olde Ice Cream Shoppe</h1>

<p>We sell amazing flavors of ice cream.</p>

<h2>Most Popular Flavors</h2>

<ul>
<?php
    foreach ($flavors as $flavor) {
        echo "<li>" . $flavor["name"] ." "."+"."likes". "</li>";
    }
?>
</ul>

</body> </html> "

Yasemin Sipahioglu
Yasemin Sipahioglu
Courses Plus Student 397 Points

names of flavors are visible but the likes

3 Answers

Austin Davis
Austin Davis
23,939 Points
Austin Davis
Austin Davis
23,939 Points

Based on your code I'm assuming you're getting an error stating that the number of likes aren't being displayed.-- To display the number of likes follow the same logic your using to display the flavor name. In your code you are using the key name via $flavor["name"] to get the value that is the flavor name...and should have a similar statement to get the likes. If you need to- click on the tab at the top to look at flavors.php and consider what key is needed to access value for "likes".

Kazimierz Matan
Kazimierz Matan
13,257 Points
Kazimierz Matan
Kazimierz Matan
13,257 Points

How about something like that:

echo "<li>" . $flavor["name"] ." +". $flavor["????"]. "</li>";

Put proper value instead of ???? (check in model).

Yasemin Sipahioglu
PLUS
Yasemin Sipahioglu
Courses Plus Student 397 Points
Yasemin Sipahioglu
Yasemin Sipahioglu
Courses Plus Student 397 Points

I got that I forgot the array, it is so good to get answer very descriptive, thanx Austin and Kazimierz

I just need $flavors ["likes"]