How Can I Avoid Global Variables When Accessing a Database in a PHP Class?

Avoid using global variables when accessing the database in PHP classes

Your code encountered an error while trying to access a global variable pagi in a $db class method. This error indicates that the query() method was called on a non-object, which indicates that the $db variable is not available in the class.

Dependency Injection Solution

The recommended way to solve this problem is to use dependency injection and pass the database object as a parameter into the class instead of relying on global variables. An example is as follows:

<code class="language-php">$db = new DB_MySQL("localhost", "root", "", "test");
include_once("pagi.php");

$pagination = new Paginator($db);
$records = $pagination->get_records("SELECT * FROM `table`");</code>

<code class="language-php">class Paginator
{
    private $db;

    public function __construct(DB_MySQL $db)
    {
        $this->db = $db;
    }

    public function get_records($q)
    {
        $x = $this->db->query($q);
        return $this->db->fetch($x);
    }
}</code>

In this example, we create a new class called Paginator that passes in the DB_MySQL object as a dependency through the constructor. This injects the database instance into the class, making it available to methods within it.

Advantages of Dependency Injection

Using dependency injection has the following advantages:

• Decoupling: It decouples a class from its direct dependency on the DB_MySQL class, making it more flexible and easier to test.
• Testability: Unit testing becomes easier because you can mock dependencies and isolate test classes.
• Avoid hidden dependencies: Unlike using the global keyword, dependency injection makes it clear that the class depends on a specific dependency.

Other methods

Another way to access the database object is to inject it directly into the method that requires it, like this:

<code class="language-php">$db = new DB_MySQL("localhost", "root", "", "test");
include_once("pagi.php");

$pagination = new Paginator();
$records = $pagination->get_records("SELECT * FROM `table`", $db);</code>

<code class="language-php">class Paginator
{
    public function get_records($q, DB_MySQL $db)
    {
        $x = $db->query($q);
        return $db->fetch($x);
    }
}</code>

However, dependency injection via constructors is generally preferred as it provides better encapsulation and testability.

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