How to use $db from another .php to another .php's class using OOP?

Question

I am new to OOP PHP, and I've been working with procedural API since I've started Web Development so I'm having a hard time migrating to OOP.

so let's say I have this four .php file and structures below.

connection.db.php

<?php
    define("DB_HOST", "127.0.0.1");
    define("DB_USER", "root");
    define("DB_PASS", "");
    define("DB_NAME", "sample_db");

    $db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

    echo (!$db->connect_error) ? NULL : die("<pre>Unable to connect to the MySQL Server -> $db->connect_error</pre>");
?>

sampleclass.class.php

<?php
    public $db;

    class MySQLqueries {
        public function samplefunction($queryString) {
            $sqlQry = mysqli->query($queryString);

            return ($sqlQry) ? "<pre>Query Executed Successfully</pre>" : die("<pre>An error occured -> $db->error</pre>");
        }
    }
?>

includes.inc.php

<?php
    error_reporting(0);
    date_default_timezone_set("Asia/Manila");

    require 'connection.db.php';
    require 'sampleclass.class.php';
?>

index.php

<?php
    require 'includes.inc.php';

    $todo = new MySQLqueries;
    echo $todo->samplefunction("SELECT `sample_column` FROM `sample_table` WHERE `sample_column` = 'sample_value';");
?>

As you may or may have not noticed, my problem is how to use the $db from connection.db.php in the samplefunction of sampleclass.class.php

You may ask me, why not create a __construct() method in sampleclass.class.php and just move the "$db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);" there ?. well if I'm to do this, all other classes will have to have it's own constructor and $db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME); declared there right?

With $db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME); on the connection.db.php I assume that I can reuse it versatile-ly.

You see I've declared a public $db; in sampleclass.class.php but I don't know what's next. If someone can help me dearly, then I would be a lot grateful.

PS: I've already consulted Dr. Google Ph.D. regarding this matter, and I only got tons of articles or video tutorials showing how to use and do OOP PHP-MySQLi Programming but without using classes and methods.

Answer 1

You'd be best to create a DB class or harnessing an already created one to achieve what you're trying to do.

The usual flow for things like this is call Lazy Loading/Dependency Injection. Where you're passing the required objects into the class.

^{If you choose this path, I suggest that you read up on Dependency Injection, as many things do, it has pros AND cons but is essential in OOP.}

As Ben Stated in the comments:

Dependency injection is a key principle in OOP. I'd like to add that if you are serious about OOP you should also look into autoloaders in general, PSR-4 and Composer.

A side not on the above mentioned, you'd be best to look at PHPTheRightWay, they list a lot of stuff, including Dependency Injection.

You'll end up creating something like. It'd be better if you followed this example to understand how it works:

Class DB {

    function __construct($host, $user, $pass, $db) { 
        return $this->connect($host, $user, $pass, $db); 
    }

    function connect($host, $user, $pass, $db) {
        //..connect and all.
    }

    //...the rest of your functions/class...
}

^{You can construct this anyway you please. Or just make the mysqli object accessible, allowing you to call it.}

Now we get to the fun stuff. Actually injecting it into your class;

Class Foo {

    $private $db;

    // your construct method here will ONLY except a `DB` class instance/object as $db. 
    // Try it with anything else and learn from the errors to understand what I mean.
    function __construct(DB $db){
        $this->db = $db;
    }

}

$db = new DB($host, $user, $pass, $db);
// you can error check it here

$foo = new Foo($db);// inject the $db object.

---

If you just want to share the resource, you could harness global, but it is strongly discouraged.

include('connection.db.php');

class MySQLqueries {
        public function samplefunction($queryString) {
            global $db;
            $sqlQry = mysqli->query($queryString);

            return ($sqlQry) ? "<pre>Query Executed Successfully</pre>" : die("<pre>An error occured -> $db->error</pre>");
        }
}

^{If you chose this path, you'd be best to assign the global instance of $db to an internal class variable, like $this->db.}

Answer 2

As you're alread using objects, you probably should also go to a OOP (Object Oriented Programing) implementation.

So, you may put your $db variable as a static variable in the sampleclass.class.php file, as in:

class MysqlConn
{
    public static $db;
}

Like that you instantiate the mysqli object using this code:

MysqlConn::$db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

Even better, you could create a method to instantiate the $db connection, initialize the variables in that class:

<?php
class MysqlConn
{
    protected
        static $DB_HOST = "127.0.0.1",
        static $DB_USER = "root",
        static $DB_PASS = "",
        static $DB_NAME = "sample_db";

    protected static $db;

    public static function getDB()
    {
        if (!isset(self::$db) {
            self::$db = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

            if ($db->connect_error) {
                die("<pre>Unable to connect to the MySQL Server -> $db->connect_error</pre>");
            }
        }
        return self::$db;
    }
}

MysqlConn::getDB();

In index.php, you may get the db instance with the following code, not reconnecting every time:

$db = MysqlConn::getDB();