I have a package that implements tagging functionality. In order to create new tags, I need to access the model inside the vendor folder. However, I get an error when I try the following:
$tag = new vendor\codecourse\taggy\src\models\tag;
Class 'vendor\codecourse\taggy\src\models\tag' not found
How do I access this model?
vendor is assumed. Look at the namespace declaration in the file you want to use to figure out the rest. Per https://github.com/codecourse/taggy/blob/master/src/Models/Tag.php, it'll be:
$tag = new Codecourse\Taggy\Models\Tag;
Related
My apologies if this exists already but my search-fu can not find the answer.
I have a composer package, and want to use my model ONLY IF an existing model doesn't exist (or extend the custom model), but I can't seem to figure out how to specify the "use" command properly inside my composer model. Since I won't know the name of the "app" using the package, I can't extend it.
<?php
namespace MyComposer\Package\Models;
use Illuminate\Database\Eloquent\Model;
class MyPackageModel extends Model
{
If I put it as a config option, I can't use that in the extends i.e class MyPackageModel extends config('custom_model_name')
I had thought I should do the check in the ServiceProvider but I can't seem to find the right code to register the proper model name to use in there.
Thanks.
I've done something similar to this before, I believe. But my approach was slightly different. See if it makes sense:
Create a base class in your own package. This will be the fallback
model which will be used if the "local" package (the one consuming
your package) doesn't have it's own version of it;
Create a config file which states which model will be used. The default is the model inside your own package (i.e. the fallback);
After installing and setting up your package, if a user does nothing they will automatically have your base model available. If they wish to override your base model with a custom local version, they can simply extend your base model and alter the model to be used in their config file.
I've also found that sometimes it's useful for the base model to 1) implement an interface that can be checked in your package's logic without relying on a specific class (which, after all, is meant to be overridden, right?); and 2) have most of it's logic inside a trait which the "local" model can use without ever having to extend your model (crucial if the local model already extends some other class for whatever reason).
How you approach the code would very much depend what you plan to do with that model. Say, for example, you have a supporting class that creates media entries in your database. Here's your packages model:
<?php
namespace Namespace\Package;
class Media
{
//...
}
And here's the default config:
<?php
return [
'model' => \Namespace\Package\Media::class,
];
And here's a sample manipulation, where you actually account for the local app to override your own model:
<?php
namespace Namespace\Package;
class MediaManager
{
protected function getModel()
{
$model = config('package.model');
return new $model;
}
public function createMedia($attributes = [])
{
$media = $this->getModel($attributes);
$media->save();
return $media;
}
}
That is to say, you never reference any Media model literally. You do your manipulations via the MediaManager. Of course the logic is very simplistic, but hopefully it's enough to get the bigger picture.
This question already has answers here:
Troubleshooting referencing a model in a laravel controller
(2 answers)
Closed 6 years ago.
I'm finding the distinction between "Controller" and "Model" in laravel 5.2 to be very blurry.
I use artisan to create a RESTful controller, and in the store method, I try to create a new object.
// store in the database
$car = new App\Models\CarModel;
Then I get the error as follows:
Class 'carfreak\Http\Controllers\App\Models\CarModel' not found
So it all seems to come down to the namespace of the controller, but I don't understand it.
The name space describes the controller, right?
So why is my reference the model, being built on the controllers path? It shouldn't have anything to do with it... right?
EDIT: After trying various suggestions, I've concluded there are three things to look at:
Each class has a namespace set, correctly describing the folder where the class is located
In the controller, have the statement "Use app\models\CarModel"
refer to the model in the controller.
Each seems to be correct, but I still get the error that it cannot find the model
This is a namespace problem in php.
You just use like this.
$car = new \App\Models\CarModel;
or
use App\Models\CarModel;
....
class {
$car = new CarModel;
}
First of all check name space in Model file , Define name space in model file
namespace App\Http\Models;
And then use
use App\Http\Models\CarModel;
Well, here it is. I solved it by asking myself this question: If I'm having so much trouble namespacing, referencing and organising my models, then maybe I can get artisan to do it for me.
The post that got me thinking differently was Mansoor Akhtar's advice here: How do I instruct artisan to save model to specific directory?
Get artisan to make the model in the right place first time.
php artisan make:model Models/CarModel
In the Controller, reference the model correctly
use name-of-app\Models\CarModel;
There may or may not be cache flushing involved in my problem. I was eventially restarting my XAMPP after every change to ensure no caching was involved. I also tried
php artisan cache:clear
You'll need to add a use statement to the top of your class.
Try:
use carfreak\Models\CarModel;
class ...
This assumes that your model is in the carfreak\Models namespace, and in a Models folder within your App / carfreak folder, otherwise you'll just need use carfreak\CarModel;.
I believe you have just ran the artisan command to create the model and you didn't move the CarModel file to Models folder. (Correct me if I'm wrong)
So in your controller add this before class declaration:
use carfreak\CarModel;
Then anywhere in your controller you can access the model like this:
$car = new CarModel;
I'd like to get the path to a package public directory (css etc) based on the package alias.
Is there anything already built into the laravel framework?
In other words something like:
public_path('myalias');
When I'm talking about alias, you would typically "alias" a module by adding the following within your service provider's boot method:
$this->package('namespace/package','alias_name');
For those wondering why someone might want to do this:
We are running a multi domain/subdomain application that makes use of a central piece of code for all of the domains and then specific packages per domain (I'll refer to them as funnels).
Each funnel has its own controllers that can possibly extend base controllers to implement their own functionality while re-using code where they can. They also have their own views.
The funnel refers to its own views by way of something like:
View::make('funnel::path.to.view')
The way we accomplish this is by doing some business logic on page load to only load the FunnelServiceProvider related to that particular domain and aliasing it to "funnel". This way our base controllers can also refer to funnel and not be tied to a particular packages views,includes,blocks etc.
My hope is to do something similar on the views so that I can simply call something like get_funnel_path() to get the path to the funnel that is currently being loaded.
The value could then be used to load css,js,images etc without worrying about the funnel path.
This would allow us to simply copy and paste views from one domain to the next without having to modify all of the paths in potentially multiple files. We could also make use of globally included files in all/most of the views.
An example of this might be the head. The head section should be the same for 99% of the files, however the path where it loads its resources should change based on the funnel.
We use the same naming conventions for css files as well as use sass, imports, merging for all of the funnels; so only the path needs to change.
You can do something like this although it will only work with your own packages and require a bit of work. Because the alias is not really stored somewhere you can easily access you have to do that yourself.
First create some kind of class to store your package names in. I called mine PackageManager:
class PackageManager {
private $packages = array();
public function addPackage($fullName, $alias){
$this->packages[$alias] = $fullName;
return $this;
}
public function getPublicPath($alias){
if(!isset($this->packages[$alias])) return public_path();
$path = 'packages/' . $this->packages[$alias];
return public_path($path);
}
}
Now let's register that class as a singleton in a service provider:
$this->app->singleton('packagemanager', function(){
return new PackageManager();
});
Then, in every package you want to register, add this call in the boot method right next to $this->package():
$this->app['packagemanager']->addPackage('vendor/package', 'alias');
After that you can do this anywhere in your application:
app('packagemanager')->getPublicPath('alias');
If you want a shorter syntax, add this helper function somewhere:
function public_package_path($alias){
return app('packagemanager')->getPublicPath($alias);
}
And just do:
public_package_path('alias');
Am having problems understanding where classes should be kept in CI. I am building an application that describes / markets mobile phones.
I would like for all of my functions (i.e. getphone, getdetails etc.) to reside in one class called Mobile - I understand that this file should be called Mobile.php and reside in the controllers folder.
Can I then have multiple functions inside Mobile.php? E.g.
public function getphone() {
xxx
xx
xx
}
public function getdetails() {
xxx
xx
xx
}
Or do I need to put each function in its own class?
I'd really appreciate looking at some sample code that works. I've been going through the documentation and google for a few hours, and tried all sorts of variations in the URL to find a test class, but without much luck! I've even messed around with the routes and .htaccess...
All I am trying to achieve is the following:
http:///model/HTC-Desire/ to be re-routed to a function that accepts HTC-Desire as a parameter (as I need it for a DB lookup). The default controller works fine, but can't get anything to work thereafter.
Any ideas?
Thanks
Actually it works like this:
Controllers and Models go to their perspective folders as you know it
If you want to create functions that are not methods of an object, you must create a helper file. More info here :
http://codeigniter.com/user_guide/general/helpers.html
Now if you want to create your own datatypes (classes that don't extend Models and Controllers), you add them to the library folder. So if let's say you want to create a class "Car" you create this file:
class Car{
function __construct(){}
}
and save it in the libraries folder as car.php
To create an instance of the Car class you must do the following:
$this->load->library('car');
$my_car = new Car();
More information on libraries here:
http://codeigniter.com/user_guide/general/creating_libraries.html
Yes, you can have as many functions in a controller class as you'd like. They are accessible via the url /class/function.
You can catch parameters in the class functions, though it's not advisable.
class Mobile extends CI_Controller{
public function getPhone($phoneModel=''){
echo $phoneModel;
//echo $this->input->post('phoneModel');
}
}
http://site.com/mobile/getPhone/HTC-Rad theoretically would echo out "HTC-Rad". HOWEVER, special characters are not welcome in URL's in CI by default, so in this example you may be met with a 'Disallowed URI characters" error instead. You'd be better off passing the phone model (or any other parameters) via $_POST to the controller.
Classes can exist both as Controllers and Models, as CodeIgniter implements the MVC pattern. I recommend reading more about that to understand how your classes/functions/etc. can best be organized.
Off the top of my head, Pyro CMS is an application built with CodeIgniter and the source code is freely available. I'm sure there are others.
I think it's best you handle it from one perspective, that is; create a utility class with all your functions in it.
The answer to the question of where to put/place the class file is the "libraries" folder.
This is clearly stated in the documentation. Place your class in the libraries folder.
When we use the term “Libraries” we are normally referring to the
classes that are located in the libraries directory and described in
the Class Reference of this user guide.
You can read more on creating and using libraries Creating Libraries — CodeIgniter 3.1.10 documentation
After placing the newly created class in the libraries folder, to use just simply load the library within your controller as shown below:
$this->load->library('yourphpclassname');
If you wish to receive several arguments within you constructor you have to modify it to receive an argument which would be an array and you loading/initialization would then be slightly different as shown below:
$params = array('type' => 'large', 'color' => 'red');
$this->load->library('yourphpclassname', $params);
Then, to access any of the functions within the class simply do that as shown below:
$this->yourphpclassname->some_method();
I hope this answers your question if you have further question do leave a comment and I would do well to respond to them.
I have a user class with static methods getById and getByUsername
I have the class in the application/libraries folder
How do I call the classes from a controller?
Theory 1:
$this->user = new User();
$this->user::getById;
Theory 2:
$user = new User();
$user::getById;
or is there a clean way of doing it much like how Kohana helpers do it; much like:
text::random();
here's what I am trying to accompplish:
I want to call a static mehthod in the user library from my controller
In PHP you usually include the file (User.php) and the static methods are ready
User:getById
but then how would I do the same thing in an MVC framework?
shall I do an include too?
Like include ('User.php');?
User::getById();
and
User::getByUserName();
Edit: In response to your question edit, generally frameworks have an auto-loading mechanism that will find and load a class file for you once you reference that class. So when you type User::getById(), the PHP interpreter will see that it needs to load the User class (if it hasn't been loaded already), and run the autoloading procedure to find the correct code to include.
I've never used Kohana, but I would be quite surprised if it didn't use some form of autoloading. If it does not, then yes, a simple include('User.php') will be enough to make the static method calls to User work.
The confusing thing is Kohana's convention of writing "helper" classes with lowercase names.
Your user php file will probably all ready be loaded if your using it as a model, so you can use zombat's suggests of User::getById();.
I don't like to follow their naming convensions for helpers or libraries and instead do:
require_once(Kohana::find_file('libraries', 'user_utils', TRUE, 'php'));