如何连接到同一个类中的两个不同的MySQL DB

Question

I have this class called dataBase . Looks like this

    class dataBase
    {

        private $conexion;
        private $paisConexion;
        var $db;
        function __construct($db='default')
        {
            $this->db = $db;
            include '../settings/variables.php';

            if(isset($bbdd)){

            $conexion = mysql_connect($bbdd["server"], $pais[0]['user'], $pais[0]['pass']) or die('No se pudo conectar: '.mysql_error());
            // Seleccionamos la base de datos
            mysql_select_db($x[0]['database']) or die('No se pudo seleccionar la base de datos');

            if($conexion)
            {   
            $paisConexion = mysql_connect($bbdd["server"], $pais[$this->db]['user'], $pais[$this->db]['pass']) or die('No se pudo conectar: '.mysql_error());

            mysql_select_db($pais[$this->db]['database']) or die('No se pudo seleccionar la base de datos');

            }


            }
            else{
                echo 'El sistema no se pudo conectar a la base de datos.';
                exit;
            }
        }

public function execute($sql)
    {
        $result = mysql_query($sql) or die("ERROR: Ejecuci&oacute;n de consulta: $sql<br>
");
        return $result;
    }
}

I am trying to make two connection to two different database using the variable $conexion and $paisConexion .

My question is is it possible to do something like this .

I mean suppose I am creating an object for the class like this

$obj = new dataBase(1); $res = obj->execute($sql);

So how the the class will decide which of the connection it has to use ? .

I think I am doing this wrong way . If any one has any idea please let me know

Thanks in Advance

Answer 1

You can't create one class for both databases. Unless you pass some parameter that specifies witch connection to use. Also than you must use two different variables for different connections. And don't use deprecated mysql_* functions

class DataBase {
    // only private variables accessed by functions
    private $localDb, $remoteDb;
    // Always use constants instead of magic numbers
    const LOCAL = 1, REMOTE = 2

    public function _construct() {
        $this->localDb= new PDO('mysql:host=localhost;dbname=test', 'username', 'password');
        $this->remoteDb= new PDO('mysql:host=remore;dbname=test2', 'username', 'password');
    }

    // Can't use constants in function header           - - - - -v
    public function execute($queryString, $params = [], $useDb = 1) {
        // static:: will take variable from this class and not from parent class (if extends something)
        if ($useDb == static::LOCAL) {
            $db = $this->local;
        } elseif ($useDb == static::REMOTE) {
            $db = $this->remote;
        }

        $query = $db->prepare($queryString);

        // Usage of prepared statement
        return $query->execute($params);
    }
}

$db = new DataBase();
$db->execute(
    'SELECT * FROM table WHERE column = :columnVal', // named placeholders instead of tons of '?' 
    [':columnVal' => 5], // some parameters
    DataBase::LOCAL // Constant from class
);