Hjælp til en database for denne kode

Hej allesammen!

Jeg sider med denne counter, Som køre på en MySql db, Problemmet er at jeg ikke har en DB til den, ville det være muligt at lave en ud fra denne kode....Vis (JA) hvordan ????


>>>>>>>>>>KODEN<<<<<<<<<<


<?php
    function subtractDays( $days, $timestamp )
    {
        $time = $timestamp - ( ( $days ) * 24 * 60 * 60 );
        return date( "Y-m-d" , $time );
    }

    $besoegendeSamlet = 0;
    $besoegendeDag    = 0;
    $besoegendeUge    = 0;
    $besoegendeMaaned = 0;
    $besoegendePrDag  = 0;

    $query = "SELECT count( * ) as visits FROM visitor";
    $result = mysql_query( $query );
    $row = mysql_fetch_array( $result );

    $besoegendeSamlet = $row["visits"];

    $now = getdate();
    $now = mktime( $now["hours"], $now["minutes"], $now["seconds"], $now["mon"], $now["mday"], $now["year"] );

    $query = "SELECT count( * ) as visits FROM visitor WHERE dato BETWEEN '" . subtractDays( 0, $now ) . " 00:00:00' AND now()";
    $result = mysql_query( $query );
    $row = mysql_fetch_array( $result );

    $besoegendeDag = $row["visits"];

    $query = "SELECT count( * ) as visits FROM visitor WHERE dato BETWEEN '" . subtractDays( 7, $now ) . " 00:00:00' AND '" . subtractDays( 1, $now ) . " 23:59:59'";
    $result = mysql_query( $query );
    $row = mysql_fetch_array( $result );

    $besoegendeUge = $row["visits"];

    $query = "SELECT count( * ) as visits FROM visitor WHERE dato BETWEEN '" . subtractDays( 30, $now ) . " 00:00:00' AND '" . subtractDays( 1, $now ) . " 23:59:59'";
    $result = mysql_query( $query );
    $row = mysql_fetch_array( $result );

    $besoegendeMaaned = $row["visits"];

    $query = "SELECT TO_DAYS( now() ) - TO_DAYS( min( dato ) ) as days From visitor";
    $result = mysql_query( $query );
    $row = mysql_fetch_array( $result );
   
    $besoegendePrDag = round( $besoegendeSamlet / ( $row["days"] + 1 ), 1 );
   
    $query = "SELECT count(*) AS antal FROM onlineusers";
    $result = mysql_query( $query );
    $row = mysql_fetch_array( $result );
   
    $onlineusers = $row["antal"];
   
    $ownvisits = 1;
    if( isset( $_COOKIE["ownvisits"] ) )
        $ownvisits = $_COOKIE["ownvisits"];