php如何扩展与SQLite db连接的类

我想用php sqlite创建一个简单的web服务。我找到了一个很好的起点示例，但是我需要一个帮助来修改从mysql到sqlite的连接。我对php已经生疏了。 那么，如何更改下面发布的代码以反映sqlite db呢

class API extends REST
{
public $data = "";
const DB_SERVER = "localhost";
const DB_USER = "Database_Username";
const DB_PASSWORD = "Database_Password";
const DB = "Database_Name";

private $db = NULL;

public function __construct()
{
parent::__construct();// Init parent contructor
$this->dbConnect();// Initiate Database connection
}

//Database connection
private function dbConnect()
{
$this->db = mysql_connect(self::DB_SERVER,self::DB_USER,self::DB_PASSWORD);
if($this->db)
mysql_select_db(self::DB,$this->db);
}  

---

在必要的情况下，这就是如何进行SQLite连接

<?php
require_once("Rest.inc.php");

class API extends REST {
//inizio scop   
    public $data = "";

    public function __construct(){
        parent::__construct();              // Init parent contructor
        $this->dbConnect();                 // Initiate Database connection
    }

    public function processApi(){
//  echo "HELLO API";
        $func = strtolower(trim(str_replace("/","",$_REQUEST['rquest'])));
        if((int)method_exists($this,$func) > 0)
            $this->$func();
        else
            $this->response('',404); // If the method not exist with in this class, response would be "Page not found".
    }

    private function dbConnect(){
    //  echo "HELLO BOY";
        try {
    $this->db = new PDO('sqlite:model.sqlite3');
    $this->db->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
    }
    catch( PDOException $e ) {
    echo $e->getMessage();
    }
    return $this->db;

    }
    ///
    private function discipline(){

    echo $this->get_request_method();
    if($this->get_request_method() != "GET")
    {
    $this->response('',406);
    }
    $risultato = array();   
    $result = $this->db->query('SELECT * FROM discipline' , PDO::FETCH_ASSOC);

    $rows = count ($result);    
    if ($rows >0) {

    //echo "Number of righe: $rows";
    foreach($result as $row)
    {
    $risultato[] = $row;
    }
    $carso = $risultato;
    echo json_encode($carso, JSON_PRETTY_PRINT);
    $this->response($this->json($result), 200);
    $data = $carso;
    } else {
    $this->response('',204); // If no records "No Content" status;
    } 

    }

    public function json($data){
        if(is_array($data)){
            return json_encode($data, JSON_PRETTY_PRINT);
        }
    }

}   

// Initiate Library

$api = new API;
$api->processApi();
?>

是否尝试查找mysql函数的SQLite版本？SQLite是一个嵌入式数据库，所以它既不使用服务器，也不使用用户/密码。因此，您可以100%删除显示的代码。