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PHP Build a Basic PHP Website (2018) Adding a Basic Form Forms and User Input

James Barrett
James Barrett
13,253 Points
Posted by James Barrett
James Barrett
13,253 Points

Can somebody walk me through what this code does? I was unclear in the pevious section of PHP basics too.

<?php

$flavors = array();

$flavors[1] = "Cake Batter";

$flavors[2] = "Cookie Dough";

if (isset($_GET["id"])) {

    if (isset($flavors[$_GET["id"]])) {

        echo $flavors[$_GET["id"]];

    } else {

        echo "B";

   }

} else {

    echo "A";

}

?>

2 Answers

Codin - Codesmite
Codin - Codesmite
8,600 Points
Codin - Codesmite
Codin - Codesmite
8,600 Points

In the following example we assume the url is "www.website.co.uk/index.php?id=1", if for example it was "index.php?id=2" the value of $_GET["id"] would be 2 instead of 1.

<?php

$flavors = array();  // Initialize empty array $flavors

$flavors[1] = "Cake Batter"; // Push "Cake Batter" to $flavors with key index 1

$flavors[2] = "Cookie Dough" // Push "Cookie Dough" to $flavors with key index 2

if (isset($_GET["id"])) {  // If $_GET["id"] is set, $_GET[] is an associative array of variables stored in the url for example "www.website.co.uk/index.php?id=1"

    if (isset($flavors[$_GET["id"]])) { // if $flavors[1] is set if for example the url was "www.website.co.uk/index.php?id=1"

        echo $flavors[$_GET["id"]]; // echo $flavors[1] if for example the url was "www.website.co.uk/index.php?id=1"

    } else { // else if $flavors[1] is not set if for example the url was "www.website.co.uk/index.php?id=1"

        echo "B"; // echo "B"

   }

} else { // else if $_GET["id"] is not set

    echo "A"; //echo "A"

}

?>
Matthew Bilz
Matthew Bilz
15,829 Points
Matthew Bilz
Matthew Bilz
15,829 Points

Part one of this:

If the URL has a query string with one of the ice cream flavors, i.e. 'www.flavors.php?id=1', or the $_GET variable has been passed along to the page, then the code would execute the statement inside of the if(isset($_GET["id"])) bracket.

If not, then you would echo 'A' at the bottom of the code.

Then, if the id in the query string is either 1 or 2, then it would execute the next section of the code inside the if(isset($flavors[$_GET["id"]])), which would echo out the flavor associated with the $flavor with the id of 1 or 2, depending on the query string's value (if it's 1 or 2).

If not, then the code would echo out "b" because there is a query string, but the number is neither 1 nor 2, so there is no flavor array associated with the query id.

I hope this helps at all!