PHP class scope

Question

I have several classes in an application that I am currently building, and I want to have one access some of the other's member functions but i can't seem to do it.

The first class is called MySQLDB:

class MySQLDB{
public $connection;

function __construct(){
//connects to database
}

function login($username, $password){
//queries database...
}
}

Then I have a class called Session:

class Session{
//variables
//constructor

function processlogin($username, $password){
$database->login($username, $password);
}

Then after this I have two class declarations:

$database = new MySQLDB();
$session = new Session();

No matter where i put these statements in relation to the classes I still get the same error:

PHP Notice:  Undefined variable: database in C:\inetpub\wwwroot\cmu\include\session.php on line 52
PHP Fatal error:  Call to a member function login() on a non-object in C:\inetpub\wwwroot\cmu\include\session.php on line 52

I have seen some suggestions that would suggest putting the new database object inside the Session class declaration but I want to avoid doing so because I use the database class several other places in the code and I don't want to open up multiple connections to the database.

Sascha Galley · Accepted Answer · 2011-05-30 19:27:15Z

4

since you want to have acces on a globally set variable, you can either gain access to it with global:

function processlogin($username, $password){
    global $database;
    $database->login($username, $password);
}

or use the variable as a parameter for the contructor and remember the database object reference in the class Session:

class Session{

    private $database;

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

    function processlogin($username, $password){
        $this->database->login($username, $password);
    }

}

and then you call:

$database = new MySQLDB();
$session = new Session($database);

this comes in handy, if you use more functions afterwords, that also need access to the database object.

edited May 30, 2011 at 19:27

answered May 30, 2011 at 19:20

Sascha Galley

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3 Comments

Drew Galbraith Over a year ago

Thanks I just used the global $database statement and it worked just fine for now. I plan on trying the second method when I get a chance but I want to check with you to make sure that it references the object and does not create a duplicate.

gion_13 Over a year ago

I needed smth similar for my latest project and I realized I was running about 10 instances of the db for the same thing. It may be a problem if the MySQLDB instance needs to be used by other classes simultaneously, because the $database object in the Session instance will not refer to the same instance. So if you alter the object from another place, that modification will not be visible to your class. Say you want to do smth like : $database->setEncoding('utf-8');; then you will have to do the exact same thing for your Session db : $session->database->setEncoding('utf-8')

RiggsFolly Over a year ago

Down use global in a class. it totally destoys the encapsulation of your class. Only use method 2 in this answer

gion_13 · Accepted Answer · 2018-01-18 14:13:56Z

4

you could pass a reference to the MySQLDB instance in the Session constructor.

class Session{
    public $db;

    function __construct(&$db=null){
        if($db == null)
            $this->db = new MySQLDB();
        else
            $this->db = $db;
    }
    // ....
}

$database = new MySQLDB();
$session = new Session($database);

edited Jan 18, 2018 at 14:13

answered May 30, 2011 at 19:24

gion_13

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Comments

Tomasz Kowalczyk · Accepted Answer · 2011-05-30 19:16:32Z

2

You are creating two global variables. If you're doing such thing, you need to declare variables you want to use in function with "global" keyword:

function processlogin($username, $password){
global $database;
$database->login($username, $password);
}

Despite this will work, I highly recommend reading about Dependecy Injection mechanism in which you'd pass $database variable as a parameter to processlogin() method, or set it as a private member of that class in constructor / setter. That way database connection will be interchangeable and you'll get more flexibility in your code.

answered May 30, 2011 at 19:16

Tomasz Kowalczyk

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Comments

Felix Kling · Accepted Answer · 2011-05-30 19:17:48Z

2

$database is not defined inside processlogin nor passed as parameter, hence the function has no access to it.

You could pass it as constructor parameter to Session:

class Session {
    private $db;

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

    public function processlogin($username, $password){
        $this->$db->login($username, $password);
    }
}

$database = new MySQLDB();
$session = new Session($database);

answered May 30, 2011 at 19:17

Felix Kling

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2 Comments

Gerald Goshorn Over a year ago

why not create a parent class that has child classes and then make a call to the parent or child class?

Felix Kling Over a year ago

@Gerald: You mean Session should be a child of MySQLDB? Inheritance expresses a is-a relationship, but a Session is not a MySQLDB. A session uses a DB, which is what composition is for. It just makes more sense semantically, but of course you can do whatever you want ;)

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