<?php
//db_class.php

error_reporting(E_ALL);


class Db_Connect {

   var $host;
   var $user;
   var $pasw;
   var $MYSQL_ERRNO;
   var $MYSQL_ERROR;
   var $query;

   function make_connect($host, $user, $pasw) {

      $this->host = $host;
      $this->user = $user;
      $this->pasw = $pasw;

      $link_id = mysql_connect($this->host, $this->user, $this->pasw);

if (!$link_id) {
die("Impossibile connettersi al server mysql " . "<br>" .
"Dettagli: " . $this->MYSQL_ERROR . "Errore No: " . $this->MYSQL_ERRNO);
}
else {
return $link_id;
}
}


   function select_db($db_name) {

if (!(mysql_select_db($db_name))) {
die("Impossibile selezionare il database " . $db_name . "<br>" .
"Dettagli: " . $this->MYSQL_ERROR . "<br>" . "Errore No: " . $this->MYSQL_ERRNO);
}
}



function sql_query($query) {


if (!(mysql_query($query))) {
die("Impossibile eseguire query al database " . "<br>" .
"Dettagli: " . $this->MYSQL_ERROR . "<br>" . "Errore No: " . $this->MYSQL_ERRNO);
}
else $result = mysql_query($query);


      return $result;
   }


function close_connection() {

       mysql_close($this->make_connect());
   }
}

class Result_Db extends Db_Connect {


function get_data() {

         $res = $this->sql_query($query);

         while($db_data = mysql_fetch_array($res)) {

               echo $db_data["tot"] . "<br>";

         }
     }
}

$obj = new Result_Db;
$obj->make_connect('localhost', 'root', '');
$obj->select_db('sottilelinearossa');
$obj->sql_query('select count(id) as tot from articoli');
$obj->get_data();

?>


I get this message because the variable is not valid when call the method sql_query in
the child class Result_Db.
How to fix this problem ?


Notice: Undefined variable: query in c:\programmi\apache group\apache\users\kioto\db_class2.php on line 66
Impossibile eseguire query al database
Dettagli:
Errore No:


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