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patternphpMinor

MySQL database connection in the constructor

Submitted by: @import:stackexchange-codereview··
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thedatabasemysqlconstructorconnection

Problem

I'm an absolute beginner in PHP OOP in search of the "Holy Grail" of connecting to MySQL database once and reusing this connection for the whole site.

classes/db.php

 ' . mysql_error());
        mysql_select_db(DATABASE, $connection) or die('Database error -> ' . mysql_error());
    }
}
?>


classes/users.php



users.php


read();
?>
';
    foreach($users as $user){
        echo '' . $user['use_name'] . '';
    }
    echo '';   
}
?>


My doubts are mostly on the Users class part:

function __construct(){
    $db = new DB();
}


It seems to be an easy way to have the connection available, but I read somewhere that instantiate the db connection in the constructor is a bad idea. Can you explain why, and if there's a better way to have the db connection easily available in every class that needs it?

I read a similar question here, but I can't understand the abstraction/holding reference/singleton suggestion from the accepted answer, and the lack of a full practical example doesn't help me.

Solution

A better practice is passing a created DB instance to the Users class. Imagine that you have a Products class, an Orders class etc. Will all of their constructors create a separate MySQL connection? It doesn't look a good idea since all of them could use the same database connection.

class Users {
    private $db;

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

    ....
}

class Products {
    private $db;

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

    ....
}

$db = new DB();
$users = new Users($db);
$products = new Products($db);

Code Snippets

class Users {
    private $db;

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

    ....
}

class Products {
    private $db;

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

    ....
}

$db = new DB();
$users = new Users($db);
$products = new Products($db);

Context

StackExchange Code Review Q#5732, answer score: 3

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