I'm developing a component for Joomla 3, but I don't understand how to use the datamodel in joomla 3. Am I supposed to use the Legacy-classes? Will these not be deprecated after a while?
I see most of the joomla components are using the class JModelList, which extends JModelLegacy. To load such a class I can use
JModelLegacy::getInstance('classname');
but is there a way to do this without using legacy-classes? I would think that the word legacy implies that the code hase been replace for something newer/better/hotter? Please enlighten me if anyone knows better ...
PS: I also asked this question on the Joomla - forum, but got no replies (http://forum.joomla.org/viewtopic.php?f=706&t=837707)
regards Jonas
I tend to use JModelList and JModelForm because this way PHP error logging will notify during code execution if your class has failed to meet some of its "contractual" class obligations from any of the parent/super classes.
The answer depends on yourself, and there is really no "wrong" answer here. Pick the method which makes the most sense to you and go with it.
** EDIT **
You can always load and instantiate a model class like so, using User's registration model as example.
JLoader::register('UsersModelRegistration', JPATH_ROOT . '/components/com_users/models/registration.php');
$model = new UsersModelRegistration();
Related
I'm developing a Laravel 5 package where I have a "Member" model which currently extends App\User model. I would like to know the best practice to let any developer use a custom "Member" model instead of the one from the package. This is for example to allow a developer use another table.
One approach that seems to work without having done a deep test with it is to make an alias in my package service provider in the register() method:
$MemberModel = 'MyVendor\MyPackage\Member';
$loader = \Illuminate\Foundation\AliasLoader::getInstance();
$loader->alias('MyMember', $MemberModel);
In this case I have to:
Replace all entries in the code from the original class to the alias
Delete all php "use" entries related to it
Get the value of $MemberModel from a config file or the database
But I don't know if it is a good way to solve it or It may cause any conflict.
Is there any other and better approach for this goal? Thanks in advance!
I finally had to test by myself this approach without haven't read the solution anywhere else, but anyway everything seems to work fine in my source code.
If anyone is looking for doing anything similar, the code example in my question works because the $MemberModel is defined with a value. If you want to get that value from a Model instance, as me, you have to add that code in the boot() method of the service provider.
I'm a beginner in joomla. I create own component and would like to use model from other component (exactly contentbuilder).
I find few different ways how to create model but my problem is that
class ContentbuilderModelEdit extends JModel
use JPATH_COMPONENT_ADMINISTRATOR in it. When i create model ContentbuilderModelEdit i get warnings in lines that using JPATH_COMPONENT_ADMINISTRATOR constant.
Is it possible to create model using that constant from other component?
Thanks for your answers
Alas no. The JPATH_COMPONENT and JPATH_COMPONENT_ADMINISTRATOR are defined constants, and cannot be changed.
Sometimes the developers do this instinctively (it's easy) without realizing the kind of limitation they're putting on other developers. You might consider contacting the developers and propose such a change; if they accept, you won; if they don't, write a sed script that performs the changes (replacing it with JPATH_SITE and JPATH_ADMINISTRATOR . '/components/com_contentbuilder', and apply it after each update.
Or, copy their model into your component and rename it if it supports it.
:)
To call a model from another component you need firstly to include the path of this model:
JModelLegacy::addIncludePath(JPATH_SITE . '/components/comp1/models', 'Comp1Model');
Secondly you have to create an instance of your model:
$model = JModelLegacy::getInstance('Model1', 'Comp1Model');
After that you should be able to use the methods of your model.
ref link
I would like to check if my assumption about codeigniter is right ?
We would normally extend a class when we are trying to include more functionality to the core, such as MY_Controller extends Controller, MY_Model extends Model etc...
But for example, if we are in the checkout library retrieving some checkout info(eg, product_id), we can just $this->load->library('product_lib',array('product_id'=>$product_id)) and we can easily $this->product_lib->product_name etc... from the checkout library right?
The $this->load thing is kind of equivalent to "hard code" checkout library to extend product_lib(class checkout_lib extends product_lib) to be able to use whatever methods/variables there is in the product_lib.
Please enlighten me.
In CodeIgniter $this->load is like having a resource manager (e.g. resourceManager->load("path/to/file")) and it takes care of loading the library, and passing any arguments you specify and such, easily allowing you to quickly get to using it.
So if you have a variable named product_name in your product_lib then yes calling $this->product_lib->product_name will be accessing that variable.
Really it just places the library into an array with the library name as the key and the instance of the library as the value so calling $this->product_lib is really calling something similar to $loadedLibraries['product_lib'] and returning the instance.
I hope that answers what you are asking, I'm quite tired and could have miss understood you question.
I think you misunderstood the OO paradigm and the way CI work.
$this->load is same with instantiate an object of the library/model, or load the helper file. CI have some sort of management to see if the helper/library/model already uploaded or not.
In other hand, the extends is used when defining a class, to tell PHP that the class will be inherit the parent class properties and method. A class is a blue print of object it will produce.
Maybe you can start by understanding the OO concept first. You can read this as a start, and see the reference used there.
I'm updating an php web application that is becoming multilingual, based on the Zend MVC framework, and I'm trying to figure out the best approach to passing the translation object to different classes/layers.
Most of my translation is done at the View level, but there are a few cases where I need to return status messages from custom libraries.
I could just create a property for the library and set the translator, but I'm wondering if there is a better way to integrate a translator object into an existing application?
Hold the users lanaguage in a Memento, and pass that through the program logic, when you need to do that translation use it identify the language.
If using Zend_Translate, it's best option to use register.
Zend_Registry::set('Zend_Translate', $translate);
This way all classes can find it automatically (Zend_Form, Zend_Validate, ...)
You could always instantiate the translator in the bootstrap.php so it is available to all classes as a global. That's how I'd do it since you probably need to use it everywhere. It isn't elegant, but it keeps you from having to add code wherever a class needs to throw an exception or return an error message.
If you don't have that many controllers setup can you not extend the base controller and instantiate the translator there? It should be available for use throughout the system then.
Something like this to extend:
<?php
class BaseController extends Zend_Controller_Action
{
public function init()
{
//setup translation object
}
}
?>
You might want to consider to use a dependency injector container, where the translator is an entry that you pass to the objects you need, without manually constructing the object. That way you can easily test and make more high quality (and testable) code
See other question here
How to use dependency injection in Zend Framework?
or this article about plugging ZF 2 DI into ZF1
http://adam.lundrigan.ca/2011/06/using-zenddi-in-zf1/
What are the best practices for implementing models in the MVC pattern. Specifically, if I have "Users" do I need to implement 2 classes. One to manage all the users and one to manage a single user. So something like "Users" and "User"?
I'm writing a Zend Framework app in php but this is more a general question.
The model should be driven by the needs of the problem. So if you need to handle multiple users, then a class representing a collection of Users might be appropriate, yes. However, if you don't need it, don't write it! You may find that a simple array of User objects is sufficient for your purposes.
That's going to be application and MVC implementation specific. You might well define a class collecting logically related classes, or you could define a static register on the user class. This is more of an OO question than MVC.
I'll second Giraffe by saying the use of included collections is almost always better than trying to write your own.
But I think your original question could be reworded a little differently... "Do I need a separate class to manage users other than the User class?
I use a static factory class to build all of my users and save them back to the database again. I'm of the opinion that your model classes need to be as dumbed down as possible and that you use heavy controller classes to do all of the work to the model classes.