Question

Problem: I have 2 classes, DB class and a User class, that will work if placed in the same class but when i seperate them i cant figure out how to get the user class to connect using the DB class.

I have ‘DBinterface’ class and a ‘user’ class.

***dbinterface.php

<? class dbinterface {
    var $dbHost,
        $dbUser,
        $dbName,
        $dbPass,
        $dbUserTable;


    function User() {
        $this->dbHost = 'host':
        $this->dbUser = 'user';
        $this->dbName = 'name';
        $this->dbPass = 'pass';
        $this->dbUserTable = 'table';
    }

} // End dbinterface class definition ?> 

***user.php

<?
include('dbinterface.php');
    class User {

        var $userID,
            $userName,
            $userPassword;

        function registerUser($userName, $userPassword) {
            // Connect to database
            $dbLink = mysql_connect($this->dbHost, $this->dbUser, $this->dbPass);
            if(!$dbLink) die("Could not connect to database. " . mysql_error());

            // Select database
            mysql_select_db($this->dbName);

            // Insert data
            $query = "insert into $this->dbUserTable values (NULL, \"$userName\", \"$userPassword\")";
            $result = mysql_query($query);

            // Test to make sure query worked
            if(!$result) die("Query didn't work. " . mysql_error());

            // Get the user ID
            $this->userID = mysql_insert_id();

            // Close database connection
            mysql_close($dbLink);

            // Assign the values to the data members
            $this->userName = $userName;
            $this->userPassword = $userPassword;
        } // End registerUser() ?>

PS:I have removed security and other mumbojumbo for easy legibility.
any and all help is MUCH appriciated!

Answer 1

Well. I’ll try to imagine what are you wanted to do 🙂
You need one class where you will store db variables, such as username, password and so on and 2nd class which will work with db-class params.

/* file1 */

Class db {

    private $_dbLink;
    private $_dbHost = 'host';
    private $_dbUser = 'user';
    private $_dbName = 'name';
    private $_dbPass = 'pass';
    private $_dbUserTable = 'table';

    public function connect ()
    {
        $this->_dbLink = mysql_connect($this->_dbHost, $this->_dbUser, $this->_dbPass);
        if(!$this->_dbLink) 
            throw new Exception ("Could not connect to database. " . mysql_error());
    }

    public function getLink()
    {
        return $this->_dbLink;
    }

    public function getUserName ()
    {
        return $this->dbUser;
    }

    public function getUserPass ()
    {
        return $this->_dbPass;
    }

    /* create the same methods for each variable */
}

/* file2 */

Class User {

    public function registerUser (/* params */)
    {
        $db = new db();
        $db->connect();

        /* your code */
        // Select database
        mysql_select_db($db->getDb(), $db->getLink());

        // Insert data
        $query = "insert into ".$db->getTable()." values (NULL, \"$userName\", \"$userPassword\")"; /* here is SQL-injection */
        $result = mysql_query($query, $db->getLink());
        /* and so on ...... */
    }
}

/* file3 */

// include file1
// include file2
$user = new User;
$user->registerUser(/* params */);

This is fast solution. You should learn how to create classes and project architecture. Use php.net for more information. Read about Object-oriented programming. And read about SQL-injections for safer code.