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up vote 0 down vote favorite

im verry new to classes and PDO. I've searched for hours and hours on google to find a solution for probably an easy problem.

The thing is. I have made a class to connect to my database. Then, i made a class to load information from a database using queries. Then, i use a script to combine both classes. Though, even if i connected to the database, it still has trouble understanding it in my other class. I keep getting the errors:

Undefined variable: db

Trying to get property of non-object in

Call to a member function prepare() on a non-object in

all regarding to my $query = $db->conn->prepare statement in the Invoice class.

How can i solve this??

I have the following database class:

class database
{
    private $host;
    private $dbname;
    private $username;
    private $password;
    public $conn;

    public function __construct($host,$db,$name,$pass)
    {
        $this->host=$host;
        $this->dbname=$db;
        $this->username=$name;
        $this->password=$pass;
        $dsn = 'mysql:'.$this->dbname.';'.$this->host;
        try {
            $this->conn = new PDO($dsn, $this->username, $this->password);
            $this->conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
            echo"connection made";
        } catch (PDOException $e) {
            echo"Could not connect to the database";
            die;
        }       
    }   
}   

?>

then, there is the class that builds information from a database (invoice class)

class Invoice{
    var $customerID;
    var $customerSalutation;
    var $customerFirstName;
    var $customerMiddleName;
    var $customerLastName;
    var $customerCompanyName;
    var $customerStreetName;
    var $customerHouseNumber;
    var $customerHouseNumberSuffix;
    var $customerPostalCode;
    var $customerCountry;


    function set_customer($id){
        if ((!isset($id)) or ($id=="")){
            echo "No customerID specified";
            die;
        } else {
            if (ctype_digit($id)){
                $query = $db->conn->prepare('
                SELECT CustomerID, CustomerSalutation, CustomerFirstName, CustomerMiddleName, CustomerLastName, CustomerCompanyName, CustomerStreetName,
                CustomerHouseNumber, CustomerHouseNumberSuffix, CustomerPostalCode, CustomerCountry WHERE CustomerID = :customerID');

                $array = array (
                    'customerid' => $this->customerID
                );

                $query->execute($array);                

            } else {
                echo "Invalid customer ID";             
                die;
            }
        }
    }
}
?>

I all open it, in the following basic script:

<?php
include('database.php'); //open database class
$host="localhost";
$db="dbname";
$name="root";
$pass="";

$db = new database($host,$db,$name,$pass);

include('InvoiceBuilderClass.php'); //include invoice class

$invoice = new Invoice();

$cusID="1";
$invoice->set_customer($cusID);
?>
share|improve this question
    
You never give an instance of your $db to your Invoice. How is it going to work with $db? –  Hammerstein Jul 31 at 16:46
    
Yeah i figured that was a problem, no clue how to solve it. I finaly fixed it with "global $db;" right before the query. that seemed to work. –  Stefan Oostwegel Jul 31 at 20:27

2 Answers 2

up vote 0 down vote accepted

in your class 'Invoice' add $db to the list of your member vars:

// (...)
var $customerPostalCode;
var $customerCountry;
var $db;

then you need to add a constructor to your class 'Invoice' which you pass your database-instance to:

public function __construct($db)
{
    $this->db = $db;
}

in the other "methods" (functions) of your class you need to access this member variable like this

$this->db->conn->prepare('...');

in your "basic script" you just need to pass your instance of the class "database" to your class "Invoice":

$invoice = new Invoice($db);
share|improve this answer
    
Excellent solution! Thank you for your response. I managed to fix it by adding a global $db in my function, but i guess this solution is way nicer! –  Stefan Oostwegel Jul 31 at 20:27
    
you are welcome! –  northkildonan Jul 31 at 20:29
up vote 1 down vote

$db is not visible in body of your class. You should pass it as argument (i.e. to Invoice constructor, or use global.

Get yourself familiar with variable scope (docs) in PHP.

share|improve this answer
    
Thank you for the documentation. I knew a thing or two abotu variables, but this documentation is an eye opener. Thnx :) –  Stefan Oostwegel Jul 31 at 20:28

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