How do I reference an object instance from a class initialization?

Question

Here is a small example. I want to reuse the mongodb object in many other methods in this class, without initializing it within each function.

Or is there a better approach in general?

 <?php
class Model_Users extends Model 
{

    public static function _init()
    {
        $mongodb = \Mongo_Db::instance('content');
    }


    public static function get_user($name)
    {

        $user = $mongodb->get('users', array( 'name' => $name ));

        return  $user; 
    }


}

Drew · Accepted Answer · 2012-06-20 20:45:55Z

1

Do something like this:

class Model_Users extends Model 
{
    private static $mongodb;

    public static function _init()
    {
        self::$mongodb = \Mongo_Db::instance('content');
    }


    public static function get_user($name)
    {

        $user = self::$mongodb->get('users', array( 'name' => $name ));

        return  $user; 
    }


}

This way you're creating the object for use in every method within the class.

edited Jun 20, 2012 at 20:45

answered Jun 20, 2012 at 18:59

Drew

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8 Comments

sterling Over a year ago

I tried that, but this was the response: ErrorException [ Error ]: Using $this when not in object context: $this->mongodb = \Mongo_Db::instance('content');

Drew Over a year ago

Yeah, sorry - you'll need to initialize the object instead of making a direct call to the instance method - so it will be: $this->mongodb = new Mongo_Db(); and then you can use: $this->mongodb->instance('content');

sterling Over a year ago

Thank you. That partially works, but doesnt access the instance of mongo_db that includes the relevant connection details. It essentially creates a new / fresh instance and not through the fuelphp core. I appreciate your help.

Drew Over a year ago

Is there already an variable setup for $mongodb somewhere else in the code? if so, you can use global $mongodb; and then $this->mongodb = $mongodb; - I will edit the code to reflect

sterling Over a year ago

Thank you so much. That actually didnt work, but after much trial and error I found a solution.

|

Marvo · Accepted Answer · 2012-06-20 20:23:09Z

1

How about something like this? Note, the PHP probably doesn't compile. The thrust of my post is that your model objects aren't database objects. Create a database object that is separate and provides the basic functionality that most of your model/DAO/ActiveRecord/whatever classes need. Then let your model focus on business logic, not database code. You might find that if you need to switch from Mongo to some other database later, it'll be easier because you've encapsulated all the database specific code into one place.

So, something like the following. My post is about the design, not the language.

class Db {
    private static $mongoDb;
    public static function _init() {
        self::$mongoDb = \Mongo_Db::instance( 'content' );
    }
    public static function getValue( $table, $name ) {
        $value = "";
        return self::$mongoDb->get( $table, array( "name", $name ) );
    }
}

class Model_Users extends Model {
    private $db;
    public static function _init() {
        $this->db = new Db();
    }
    public static function getUser( $name ) {
        return $this->db->getValue( "users", $name );
    }
}

answered Jun 20, 2012 at 20:23

Marvo

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2 Comments

sterling Over a year ago

Thank you so much for this excellent advice. There is a logic controller that is calling this model controller, so in effect it is doing what you are saying, but I agree that it could be abstracted even further.

Marvo Over a year ago

Glad if it helps. I wish it was a full working example, but you probably know your PHP better than I do and can make something out of it.

Scotty C. · Accepted Answer · 2012-06-20 18:58:13Z

0

I believe that is the purpose of using $this. It makes things accessible to the rest of the object.

<?php
class Model_Users extends Model 
{

    private $mongodb;

    public static function _init()
    {
        $this->mongodb = \Mongo_Db::instance('content');
    }

    public static function get_user($name)
    {

        $user = $this->mongodb->get('users', array( 'name' => $name ));

        return  $user; 
    }
}

answered Jun 20, 2012 at 18:58

Scotty C.

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1 Comment

sterling Over a year ago

I tried that, but this was the response: ErrorException [ Error ]: Using $this when not in object context: $this->mongodb = \Mongo_Db::instance('content');

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