PHP pass function name as param then call the function?

To answer your question first: No, you can't (without resorting to evilCode) pass a function name as a parameter. But: What you want to archive is a poster-child-issue for an object oriented approach using inheritance.

You'd need a base-class:

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

and your implementations :

class MyClass extends BaseClass
{
    function __construct($db) {
        parent::__contruct($db);
        $this->userDAO = DAO_DBrecord::createUserDAO($this->db);
    }
}

Just for the record: the evilCode would have been a) you could encapsulate your function in a create_function that can be used as an argument. b) you could pass the function name as a string to your function and then pass it to eval in the receiving function.

But remember: When eval or create_function looks like the answer you're probably asking the wrong questions!

See: related question

You need to do that yourself. You can use null to indicate that a default value should be used:

public function getSomething($orderBy = null, $direction = null, $limit = null) {
    // fallbacks
    if ($orderBy === null) $orderBy = 'x';
    if ($direction === null) $direction = 'DESC';

    // do something random
}

Then pass null when calling it to indicate that you want to use the defaults:

$random = $this->my_model->getSomething(null, null, 10);

Another possible solution that I use sometimes is an additional parameter at the very end of the parameter list, containing all optional parameters:

public function foo($options = array()) {
    // merge with defaults
    $options = array_merge(array(
        'orderBy'   => 'x',
        'direction' => 'DESC',
        'limit'     => null
    ), $options);

    // do stuff
}

That way you do not need to specify all optional arguments. array_merge() ensures that you are always dealing with a complete set of options. You would use it like this:

$random = $this->my_model->foo(array('limit' => 10));

It seems like there is no required parameter this particular case, but if you need one, simply add it in front of the optional ones:

public function foo($someRequiredParameter, $someOtherRequiredParameter, $options = array()) {
    // ...
}