Yii - Get data from Model - php

I have the following in my controller:
public function actionIndex() {
$userID = Yii::app()->user->getId();
$arNotifs = Notification::model()->getNotificationsByUserId($userID);
$this->render('index', array("arNotifications"=>$arNotifs, "userID"=>$userID));
}
I have the following in a file called notification.php in my models:
class Notification extends CActiveRecord {
// ------ BUNCH OF STUFF
public function getNotificationsByUserId($userId) {
$userId = (int) $userId;
$query = Yii::app()->db->createCommand();
$query->select('n.id, n.title, n.content, n.updated');
$query->from('hr4_notification_x_user nxu');
$query->join('hr4_notification n', 'nxu.notification = n.id');
$query->where('nxu.user=:userId', array(':userId' => $userId);
return $query->queryAll();
}
// ------ MORE STUFF
}
When I rem out the line
$arNotifs = Notification::model()->getNotificationsByUserId($userID);
and replace it with a static value it works fine. It seems that in my noob ways I am missing some vital step. The controller seems to have no idea what Notification is.
Thanks in advance

I believe the most elegant way to get your notifications on the controller would be something like:
$arNotifs = Yii::app()->user->model->notifications;
To achieve such, you might need to implement a getModel() method on your class that extends CWebUser. That method would return an instance of an user that extends CActiveRecord. Then your user model can have a relations() method, like the following:
class UserModel extends CActiveRecord {
public function relations() {
return array(
'notifications' => array(self::HAS_MANY, 'Notification', 'user'),
);
}
}
This will prevent you from writing that query and will make things more clear (on both, models and controller). If you will, read a bit about relations.

You cannot use the Notification model like this.
You can instantiate it with $notification = new Notification();
and then do a $notification->getNotificationsByUserId($userID);
However this would be not very good. I would move the notification code from the model to the User model.
This was you dont even need to pass the user ID.
Or maybe even better, if you make a component out of Notification and use it as a service.

Related

What is a good way of associating a Model instance with it's own Class in Laravel?

I'm sure there is a common pattern for this kind of thing, and I'm struggling with search terms to find answers, so please bear with me if is this a dupe.
I have a few Classes in my app that create pretty standard Models that are stored in a relational database, eg;
// AtsType::name examples = 'XML', 'RSS', 'SOAP'
class AtsType extends Model
{
public function ats_instances()
{
return $this->hasMany('App\AtsInstance');
}
public function import()
{
}
}
What I need that import() method to do, however, somehow invokes a class/interface/contract/whatever based upon the actual model instance. So something like this;
AtsTypeRss::import()
AtsTypeXml::import()
AtsTypeSoap::import()
I'd like them to be standalone classes, in order to eventually use some artisan commands that will generate them for a developer, along with a data migration to create the new model names into the database.
I'm just unsure how to go about this.
You could try something like (as seen here), I've searched how to use variable in namespace :
class AtsType extends Model
{
protected $import_method = 'MyMethod';
public function ats_instances()
{
return $this->hasMany('App\AtsInstance');
}
public function import()
{
$string = $this->import_method;
$class = '\\controller\\' . $string;
$newObject = new $class();
}
}

How to save database operation in laravel

Thanks for watching my first question.
I have something confused.
How could I write the operations of database into database and don't write the function in every Controller?
I have considered middleware and find that must change my route register style.
my Route is this:
Route:resource('province','\\Modules\\Info\\Controllers\\P_ProvinceController');
Dose it has some awesome methods replace this?
public function Store(Request $request)
{
$params = $request->input('data');
$params['CreateID'] = Auth::user()->id;
$params['CreateName'] = Auth::user()->name;
$params['CreateTime'] = Carbon::now();
$province = P_ProvinceModel::Create($params);
$params['Pro_Is_Del'] = 1;
$log_info['table'] = $province->getTable();
$log_info['type'] = "create";
$log_info['user'] = Auth::user()->name;
$log_info['datetime'] = Carbon::now();
LogModel::create($log_info);
if($province){
return response()->json(array(
'status' => 200,
'msg' => '新增成功',
'data' => $province
));
}else
return response()->json(array(
'status' => 500,
'msg' => '保存失败',
));
}
Thanks.
Here is how I solved across model functionality
First Create a Trait that does what you want on save.
<?php
namespace App\Models\Observers;
trait CreatedByObserver
{
public static function bootCreatedByObserver(){
/** Simply means that whenever this model is creating a model do: */
static::creating(function($model){
if(auth()->check()){
$responsiblePerson = auth()->user()->first_name . " " . auth()->user()->last_name;
} else {
$responsiblePerson = "system";
}
/** You can set any model variables within */
$model->created_by = $responsiblePerson;
});
}
}
In there do all you need to do when a record is saved/created/updated/deleted
Then In all Models you want this behaviour used add the trait.
Check them out here : https://laravel.com/docs/5.2/eloquent#events
As far i understand your question, you are asking for way to make your controller an abstract type, i.e. controller just need to handle the route and view not any other things(like database,application logic etc) which is the philosophy of the laravel Framework.
To make your controller abstract (meaning of abstract as explained aboave),
first you need to understand, "What your application logic are and what your database logic are ?"
when you understand these two things then, you can easily separate your aapplication logic and databasse logic from your controller.
For example :
For keeping your Application logic you can make service folder in your root of your project also you can make folder name 'Dao' (Database access object) in the same path of service . You need to keep these folder in autoload from your composer. Just make class for service and your Dao.
And now your application follow will be,
First Route, will hit controller then controller will need to call some method in service and then service will call the respective DAO . method.
Example :
Controller/YourController.php
Class YourController extends Controller {
public function Store(Request $request,yourservice,$yourService)
{
$this->myservice = $yourservice;
$this->myservice->store('your inputs request');
return $something ;
}
}
service/yourService.php
Class yourService {
public function store($yourinputs,yourDao $mydao){
$this->mydao = $mydao;
//you can use your application logic here
return $this->mydao->create($yourinputs);
}
And now the turn is for DAO :
dao/yourdao.php
use model // use your model here .
class yourDao {
public function create($yourdata,yourmodel $model){
$this->model = $model;
return $this->model->create($yourdata);
}
}
Now, you can see the controller just save the data in database, but don't know how it is saving the data and what are the application logic.
This explanation is just a simple way of doing project to make a controller abstract. There are other various ways of doing this. For example you can see Repository Design Pattern , which also used by laravel core .
Hope this explanation will not bore anyone . :) Happy coding .

Laravel 5 Global CRUD Class

Before anyone asks, I've looked into CRUD generators and I know all about the Laravel Resource routes, but that's not exactly what I'm pulling for here.
What I'm looking to do is create one Route with a couple parameters, and one global class that (uses/extends?) the Model controller for simple CRUD operations. We have 20 or so Models and creating a Resource Controller for each table would be more time consuming than finding a way to create a global CRUD class to handle all "api" type calls and any ajax json request like a create / update / destroy statement.
So my question is what is the cleanest and best way to structure a class to handle all CRUD requests for every Model we have without having to have a resource controller for every model? I've tried researching this and can't seem to find any links except ones to CRUD generators and links describing the laravel Resource route.
The easiest way would be to do the following:
Add a route for your resource controller:
Route::resource('crud', 'CrudController', array('except' => array('create', 'edit')));
Create your crud controller
<?php namespace App\Http\Controllers;
use Illuminate\Routing\Controller;
use App\Models\User;
use App\Models\Product;
use Input;
class CrudController extends Controller
{
const MODEL_KEY = 'model';
protected $modelsMapping = [
'user' => User::class,
'product' => Product::class
];
protected function getModel() {
$modelKey = Input::get(static::MODEL_KEY);
if (array_key_exists($modelKey, $this->modelsMapping)) {
return $this->modelsMapping[$modelKey];
}
throw new \InvalidArgumentException('Invalid model');
}
public function index()
{
$model = $this->getModel();
return $model::all();
}
public function store()
{
$model = $this->getModel();
return $model::create(array_except(Input::all(), static::MODEL_KEY));
}
public function show($id)
{
$model = $this->getModel();
return $model::findOrFail($id);
}
public function update($id)
{
$model = $this->getModel();
$object = $model::findOrFail($id);
return $object->update(array_except(Input::all(), static::MODEL_KEY));
}
public function destroy($id)
{
$model = $this->getModel();
return $model::remove($id);
}
}
Use your new controller :) You have to pass the model parameter that will contain the model key - it must be one of the allowed models in the whitelist. E.g. if you want to get a User with id=5 do
GET /crud/5?model=user
Please keep in mind that it's as simple as possible, you might need to make the code more sophisticated to match your needs.
Please also keep in mind that this code has not been tested - let me know if you see any typos or have some other issues. I'll be more than happy to get it running for you.
Unless you want to implement CRUD manually, consider to integrate a ready-made datagrid such as phpGrid.
Check out integration walkthrough: http://phpgrid.com/example/phpgrid-laravel-5-twitter-bootstrap-3-integration/ No models are required and the code is minimum. It can almost do anything.
A basic working CRUD:
// in a controller
public function index()
{
$dg = new \C_DataGrid("SELECT * FROM orders", "orderNumber", "orders");
$dg->enable_edit("FORM", "CRUD");
$dg->display(false);
$grid = $dg -> get_display(true);
return view('dashboard', ['grid' => $grid]);
}
You need one generic class for all CRUD operations and there are many ways to achieve that and one rule for all may not fit but you may try the approach that I'm going to describe now. This is an abstract idea, you need to implement it, so at first, think the URI for all CRUD operations. In this case you must follow a convention and it could be something like this:
example.com/user/{id?} // get all or one by id (if id is available in the URI)
example.com/user/create // Show an empty form
example.com/user/edit/10 // Show a form populated with User model
example.com/user/save // Create a new User
example.com/user/save/10 // Update an existing User
example.com/user/delete/10 // Delete an existing User
In ths case the user could be something else to specify the name of the model for example, example.com/product/create and keeping that on mind, you need to declare routes as given below:
Route::get('/{model}/{id?}', 'CrudController#read');
Route::get('/{model}/create', 'CrudController#create');
Route::get('/{model}/edit/{id}', 'CrudController#edit');
Route::post('/{model}/save/{id?}', 'CrudController#save');
Route::post('/{model}/delete/{id}', 'CrudController#delete');
Now, in your app\Providers\RouteServiceProvider.php file modify the boot method and make it look like this:
public function boot(Router $router)
{
$model = null;
$router->bind('model', function($modelName) use (&$model, &$router)
{
$model = app('\App\User\\'.ucfirst($modelName));
if($model)
{
if($id = $router->input('id'))
{
$model = $model->find($id);
}
return $model ?: abort(404);
}
});
parent::boot($router);
}
Then declare your CrudController as given below:
class CrudController extends Controller
{
protected $request = null;
public function __construct(Request $request)
{
$this->request = $request;
}
public function read($model)
{
return $model->exists ? $model : $model->all();
}
// Show either an empty form or a form
// populated with the given model atts
public function createOrEdit($model)
{
$classNameArray = explode('\\', get_class($model));
$className = strtolower(array_pop($classNameArray));
$view = view($className . '.form');
$view->formAction = "$className/save";
if(is_object($model) && $model->exists)
{
$view->model = $model;
$view->formAction .= "/{$model->id}";
}
return $view;
}
public function save($model)
{
// Validation required so do it
// Make sure each Model has $fillable specified
return $this->model->fill($this->request)->save();
}
public function delete($model)
{
return $this->model->delete();
}
}
Since same form is used to creating and updating a model, use something like this to create a form:
<form action="{{url($formAction)}}" method="POST">
<input
type="text"
class="form-control"
name="first_name" value="{{old('first_name', #$model->first_name)}}"
/>
<input type="Submit" value="Submit" />
{!!csrf_field()!!}
</form>
Remember that, each form should be in a directory corresponding to the model, for user add/edit, form should be in views/user/form.blade.php and for product model use views/product/form.blade.php and so on.
This will work and don't forget to add validation before saving a model and validation could be done inside the model using model events or however you want. This is just an idea but probably not the best way to it.

Eloquent models - How do i break this dependancy

Take the following:
<?php
class EqUserRepository implements IUserRepository
{
//...
public function GetUserByID( $id )
{
return User::find( $id );
}
}
You can see that it returns a User
User is extending Model (Which extends Eloquent)
class User extends Model implements AuthenticatableContract, CanResetPasswordContract
{
public function Profile()
{
return $this->hasOne( "Profile" );
}
Now imagine in our controller, we inject our concrete implementation via the interface:
class UserController extends Controller
{
private $userRepository;
public function __construct( \App\Models\Contracts\Repositories\IUserRepository $userRepo )
{
$this->userRepository = $userRepo;
All good so far!
Now we have a controller method that calls upon our injected repository and gets a User returned...
public function GetDetails( $id )
{
$user = $this->userRepository->GetUserByID( $id );
return view( 'user', array( "Model" => $user ) );
}
Now in our view we call upon the profile:
<p> Name: <?= $Model->Profile()->first_name ?> </p>
Oops, now our view is dependant on Eloquent because the Profile method on User is making calls using Eloquent.
I have been trying to figure out the best solution to break this dependancy.
I thought that having my repository map data into a User class that did not extend Eloquent was a good idea... but then you have some looping to map the classes which is a bit of a waste of time... especially considering you could just as easily use mysql_fetch_object( $query, "User" ) and have it map straight away.
So taking 1 into account, i thought the DB class must have a way of mapping direct to a type...
Have the User depend upon my IUserRepository, then my Profile function could look like this and that would certainly break the dependency, in a really good way!
.
public function Profile()
{
return $this->userRepository->GetProfileByUserID( $this->id );
}
But you ofcourse cannot DI into an eloquent model. I tried using the boot overriding method, but that doesn't like DI too, of course...
So, question time:
Can you set the return type of the DB class?
Can you somehow DI on an eloquent model?

Cannot get DataMapper to work in CodeIgniter

I'm trying to implement an ORM in a CodeIgniter application, but cannot get it to work. To start I'm just trying to instantiate a simple test model:
<?php
class Cart extends DataMapper
{
public function __construct()
{
// model constructor
parent::__construct();
}
var $validation = array(
'username' => array(
'label' => 'UserName',
'rules' => array('required', 'trim', 'unique', 'alpha_dash', 'min_length' => 1, 'max_length' => 50),
)
);
}
?>
And then in the Controller I try this:
public function __construct()
{
parent::__construct();
$this->load->model('cart');
}
public function index()
{
$cart = new Cart();
}
But I don't even get past the constructor. The debugger stops and gives me a message saying "Waiting for an incoming connection with ide key xxxxx" (random number)
BTW the cart model class file name is in lower case, but the class name in upper case. I tried both in the constructor.
I have followed the instructions for installation carefully, copying the two datamapper files to libraries and config folders, as well as autoloading the datamapper library.
But it just doesn't work. Am I missing something? The table I'm trying to map is only a test table that actually only has an id and a username field. I don't actually understand the validation array, but just followed the examples in the docs and modified to my field. The id field doesn't seem like anyone has put in the validation array.
I should also mention that I'm a newbie at CodeIgniter.
Your code seems mostly correct for use with DataMapper ORM and CodeIgniter.
To explain things a bit, DataMapper is just an abstraction layer. It handles a lot of the necessities when working with databases and mapping your objects to your tables. That being said, you don't have to load your models, etc. As long as you are autoloading your database library and datamapper library, you can use DataMapper.
The validation array lets DataMapper know the requirements to your properties. So, if you try to save an object and one of the properties that you've created/changed doesn't meet those requirements, then your save will fail and you'll get an error message:
// For example
if ($myObj->save())
{
// $myObj validation passed and is saved to db
}
else
{
// $myObj validation failed, save did not complete
echo $myObj->error->string;
}
Codeigniter already has a library named Cart, so you wouldn't want to name your model Cart. So you could rename that model to Basket or something else that makes sense.
I know you're still just trying to get things to work, but I feel you need to think about your data structure a bit. You wouldn't save the username in the Cart object, that's why we use relations. So, I would structure it a bit like this:
// baskets table (a table represents many baskets, therefore it is plural)
id
user_id
blah
blah
created
updated
// users table
id
username
email_address
created
updated
// basket model (a model represents 1 basket, therefore it is singular)
class Basket extends DataMapper
{
public function __construct()
{
parent::__construct();
}
var $has_one = array('user'); // each basket belongs to one user
var $validation = array(...);
}
// user model
class User extends DataMapper
{
public function __construct()
{
parent::__construct();
}
var $has_many = array('basket'); // each user can have many baskets
var $validation = array(...);
}
// controller
public function __construct()
{
parent::__construct();
}
public function index()
{
$basket = new Basket();
$basket->blah = 'whatever';
$basket->save();
// at this point, $basket is saved to the database
// now let's add it to the user
$user = new User();
$user->where('id', 1)->get(1);
// now we have a user
// save the relationship to the basket
$user->save($basket);
// now $basket->user_id == 1
// get the username from the basket
$u = $basket->user->get();
$username = $u->username;
// yes, there are faster and shorter ways to write most of this,
// but I think for beginners, this syntax is easier to understand
}
The CodeIgniter documentation about models states that you can load a model by calling
$this->load->model('Model_name');
in the constructor, and that you can access this model in your controller by doing
$this->Model_name->function();
So you should change your Controller code into
public function __construct()
{
parent::__construct();
$this->load->model('Cart');
}
public function index()
{
$this->Cart->functionCall();
}

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