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 .
Related
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();
}
}
I'm little confused with the database queries in the laravel, where do we have to write our queries: in controllers, models or routes ?
I've been through many tutorials and I see so much difference. Just creating confusion.
Kindly explain somebody
It depends on different factors, but in short you can write them both in Model, Controllers or Repositories
If you're writing a controller action and you need a query that you'll use only once, it's perfectly fine to write the query directly in the controller (or even in the route's closure ).
For example, if you want to get all users of type admin:
$admins = User::where('type', 'admin')->get();
Now, suppose you need to get the admins in more than one controller method; instead of rewriting the same query, you can create a Repository class to wrap the access to the users' Model and write the query inside the Repository:
class UserRepository
{
public function getAllAdmins()
{
return User::where('type', 'admin')->get();
}
}
Now in your controllers you can inject the Repository and use the same method of the Repository to get the admin users: this will keep your code DRY as you don't have to repeat the same query among the controllers' actions
Controller
public function __construct(UserRepository $userRepo)
{
$this->userRepo = $userRepo;
}
//controller action
public function index()
{
$admins = $this->userRepo->getAllAdmins();
}
Finally, let's suppose you need a query to count the number of the admin users. You could write this query in the UserRepository:
public function getAdminNum()
{
return User::where('type', 'admin')->count();
}
And it would be ok, but we can note that the User::where('type', 'admin') fragment of the query is shared with the query in getAllAdmins So we can improve this by using query scopes :
User Model
public function scopeAdmins($query)
{
return $query->where('type', 'admin');
}
by this, in the UserRepository methods we can rewrite our previous queries as:
public function getAllAdmins()
{
return User::admins()->get();
}
public function getAdminNum()
{
return User::admins()->count();
}
And i've just showed you a case in which a query would be writed inside a Model
You do not write any query in Model. Model is just for mapping the class that you are going to use for a table like User Model will be mapped to users (plural of model name).
You do not write queries in the routes closures like this
Route::get('/', ['as' => 'home', function(){
$totalProblems = Problem::count();
$solvedProblems = Problem::where('solved', 1)->get()->count();
$unsolvedProblems = Problem::where('solved', 0)->get()->count();
return view('Pages.index', ['totalProblems' => $totalProblems, 'solvedProblems' => $solvedProblems, 'unsolvedProblems' => $unsolvedProblems]);
}]);
This is considered as bad practice, its just for testing purposes.
You always write your queries in the controller method associated with your route like this
Route::get('test', 'HomeController#test');
and in your HomeController
<?php namespace App\Http\Controllers;
use App\Problem;
class HomeController extends Controller {
public function test(){
$user = User::all();
return view('Pages.test')->withUser($user); //or
return view('Pages.test')->with('user' , $user);
}
}
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.
I am using cakephp-2.x. I have one function name user_info() in the UsersController.php i want to access this in another controller name MessagesController.php
Code -
UsersController.php
public function user_info(){
$user_id=$this->Session->read('Auth.User.id');
$data=$this->User->findById($user_id);
$this->set('user_info',$data);
}
MessagesController.php
public function index(){
//$userInfo=new UsersController();
//$userInfo->user_info();
$this->user_info();
pr($data);
}
Error Message-
Fatal Error
Error: Call to undefined method MessagesController::user_info()
File: E:\xampp\htdocs\2014\myshowcam\msc\app\Controller\MessagesController.php
Line: 18
Notice: If you want to customize this error message, create app\View\Errors\fatal_error.ctp
Typically if you're trying to access a function in one controller from another controller you have a fundamental flaw in your project's logic.
But in general object usage is thus:
$otherController = new whateverMyControllerNameIs();
$otherController->functionName();
However I'm not familiar enough with cake to tell you the pitfalls of doing such a thing. For example I have no idea what this would do to routes or what other variables/objects are required to initialize a controller correctly.
EDIT:
Ref: CakePHP 2.3.8: Calling Another Controller function in CronController.php
App::import('Controller', 'Products'); // mention at top
// Instantiation // mention within cron function
$Products = new ProductsController;
// Call a method from
$Products->ControllerFunction();
Try requestAction function of cakephp
$result = $this->requestAction(array('controller' => 'users', 'action' => 'user_info'));
Why would a simple, When can complicated?
All the information for a registered user of User model is accessible in the following manner:
AppController.php
public $user_info; /* global scope */
public function beforeFilter(){
$this->user_info = $this->Auth->user(); // for access user data in any controller
$this->set('user_info_view',$this->Auth->user()); // for access user data in any view or layout
}
MessagesController.php
public function index(){
debug($this->user_info);
$my_messages = $this->Message->find('all',
array('conditions' => array('Message.user_id' => $this->user_info['id']))
}
....
layout or view.ctp
<?php echo $user_info_view['name']; ?> // email, etc
Why not take advantage of the way CakePHP handles relationships? There's a very easy way to achieve what you're trying to do without extending controllers or loading in additional controllers which seems excessive for your example.
Inside AppController's beforeFilter()
Configure::write('UserId', $this->Session->read('Auth.User.id'));
This will allow you to access the UserID from your models
Inside your User's model, create the following function
/**
* Sample query which can be expanded upon, adding fields or contains.
*
* #return array The user data if found
*/
public function findByUserId() {
$user = $this->find('first', array(
'conditions' => array(
'User.id' => Configure::read('UserId')
)
));
return $user;
}
Inside your Users controller (Minimal is better, no?)
public function user_info() {
$this->set('user', $this->User->findByUserId());
}
Inside your Messages controller
public function index() {
$this->set('user', $this->Message->User->findByUserId());
// --- Some more stuff here ---
}
And that's it, no need to be extending controllers, just make sure your Message and User model are related to each other, failing that you can bindModel or use ClassRegistry::init('User')-> for example.
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.