I'm using latest (dev-master) sonata admin and I want to create my own createAction() method for sonata admin. I have to do that, because I want to save some user information, while adding to database.
My custom controller is - S\CoreBundle\Controller\NewsAdminConroller.php
<?php
namespace S\CoreBundle\Controller;
use Sonata\AdminBundle\Controller\CRUDController as Controller;
use Symfony\Component\Security\Core\SecurityContextInterface;
class NewsAdminController extends Controller
{
/**
* Set the system user ID
*/
private function updateFields($object)
{
//some code - this is my own method
}
public function createAction(Request $request = null)
{
//code for create ... it's almost the same as default code.
}
}
Default CRUD - Sonata\AdminBundle\Controller\CRUDController.php:
class CRUDController extends Controller
{
public function createAction(Request $request = null)
{
//...
}
}
Both createAction() methods have exactly the same arguments, name ...
And it throw's me an error:
PHP Strict Standards: Declaration of S\CoreBundle\Controller\NewsAdminController::createAction() should be compatible with Sonata\AdminBundle\Controller\CRUDController::createAction(Symfony\Component\HttpFoundation\Request $request = NULL) in /home/mark/dev/project/src/S/CoreBundle/Controller/NewsAdminController.php on line 129
The Sonata\AdminBundle\Controller\CRUDController::createAction(Symfony\Component\HttpFoundation\Request $request = NULL)
Needs a Request Object, but if you don't declare it, it point to S\CoreBundle\Controller\Request
Just add
"use Symfony\Component\HttpFoundation\Request;" in top of file.
Update
Since the commit https://github.com/sonata-project/SonataAdminBundle/commit/49557c302346f57d962b83b31e2931446ff60e9c, there is no need to set the request as parameter.
The create Action is only
Sonata\AdminBundle\Controller\CRUDController::createAction()
Related
I'm building an application, now i'm created a helper
class Students{
public static function return_student_names()
{
$_only_student_first_name = array('a','b','c');
return $_only_student_first_name;
}
}
now i'm unable to do something like this in controller
namespace App\Http\Controllers;
class WelcomeController extends Controller
{
public function index()
{
return view('student/homepage');
}
public function StudentData($first_name = null)
{
/* ********** unable to perform this action *********/
$students = Student::return_student_names();
/* ********** unable to perform this action *********/
}
}
this is my helper service provider
namespace App\Providers;
use Illuminate\Support\ServiceProvider;
class HelperServiceProvider extends ServiceProvider
{
/**
* Register the service provider.
*
* #return void
*/
public function register()
{
foreach(glob(app_path().'/Helpers/*.php') as $filename){
require_once($filename);
}
}
}
i event added it as an alias in config/app.php file
'Student' => App\Helpers\Students::class,
Try putting use App\Helpers\Student; at the top of your controller beneath the namespace delcaration:
namespace App\Http\Controllers;
use App\Helpers\Student;
class WelcomeController extends Controller
{
// ...
Look more into PHP namespaces and how they are used, I believe you may have a deficient understanding about them. Their only purpose is to make so you can name and use two classes with the same name (e.g. App\Helpers\Student vs maybe App\Models\Student). If you needed to use both of those classes inside of the same source file, you can alias one of them like this:
use App\Helpers\Student;
use App\Models\Student as StudentModel;
// Will create an instance of App\Helpers\Student
$student = new Student();
// Will create an instance of App\Models\Student
$student2 = new StudentModel();
You do not need to have a service provider for this, just the normal language features. What you would need a service provider for is if you wanted to defer the construction of your Student object to the IoC:
public function register()
{
$app->bind('App\Helpers\Student', function() {
return new \App\Helpers\Student;
});
}
// ...
$student = app()->make('App\Helpers\Student');
You should never have to include or require a class file in laravel because that is one of the functions that composer provides.
You do not need a service provider to make it works. Just lets the Students class as you did:
class Students{
public static function return_student_names()
{
$_only_student_first_name = array('a','b','c');
return $_only_student_first_name;
}
}
all its methods should be static
You added the Facade correctly:
'Student' => App\Helpers\Students::class,
Finally, looks like your problem is caused by forgetting a backslash at facade name. Uses \Students instead of Students:
public function StudentData($first_name = null)
{
$students = \Student::return_student_names();
}
When using a facade, it is not necessary makes nay include, the facades were made to avoid complex includes in everywhere.
I have a BaseController that provides the foundation for most HTTP methods for my API server, e.g. the store method:
BaseController.php
/**
* Store a newly created resource in storage.
*
* #return Response
*/
public function store(Request $request)
{
$result = $this->repo->create($request);
return response()->json($result, 200);
}
I then extend on this BaseController in a more specific controller, such as the UserController, like so:
UserController.php
class UserController extends BaseController {
public function __construct(UserRepository $repo)
{
$this->repo = $repo;
}
}
This works great. However, I now want to extend UserController to inject Laravel 5's new FormRequest class, which takes care of things like validation and authentication for the User resource. I would like to do this like so, by overwriting the store method and using Laravel's type hint dependency injection for its Form Request class.
UserController.php
public function store(UserFormRequest $request)
{
return parent::store($request);
}
Where the UserFormRequest extends from Request, which itself extends from FormRequest:
UserFormRequest.php
class UserFormRequest extends Request {
/**
* Determine if the user is authorized to make this request.
*
* #return bool
*/
public function authorize()
{
return true;
}
/**
* Get the validation rules that apply to the request.
*
* #return array
*/
public function rules()
{
return [
'name' => 'required',
'email' => 'required'
];
}
}
The problem is that the BaseController requires a Illuminate\Http\Request object whereas I pass a UserFormRequest object. Therefore I get this error:
in UserController.php line 6
at HandleExceptions->handleError('2048', 'Declaration of Bloomon\Bloomapi3\Repositories\User\UserController::store() should be compatible with Bloomon\Bloomapi3\Http\Controllers\BaseController::store(Illuminate\Http\Request $request)', '/home/tom/projects/bloomon/bloomapi3/app/Repositories/User/UserController.php', '6', array('file' => '/home/tom/projects/bloomon/bloomapi3/app/Repositories/User/UserController.php')) in UserController.php line 6
So, how can I type hint inject the UserFormRequest while still adhering to the BaseController's Request requirement? I cannot force the BaseController to require a UserFormRequest, because it should work for any resource.
I could use an interface like RepositoryFormRequest in both the BaseController and the UserController, but then the problem is that Laravel no longer injects the UserFormController through its type hinting dependency injection.
In contrast to many 'real' object oriented languages, this kind of type hinting design in overridden methods is just not possible in PHP, see:
class X {}
class Y extends X {}
class A {
function a(X $x) {}
}
class B extends A {
function a(Y $y) {} // error! Methods with the same name must be compatible with the parent method, this includes the typehints
}
This produces the same kind of error as your code. I would just not put a store() method in your BaseController. If you feel that you are repeating code, consider introducing for example a service class or maybe a trait.
Using a service class
Below a solution that makes use of an extra service class. This might be overkill for your situation. But if you add more functionality to the StoringServices store() method (like validation), it could be useful. You can also add more methods to the StoringService like destroy(), update(), create(), but then you probably want to name the service differently.
class StoringService {
private $repo;
public function __construct(Repository $repo)
{
$this->repo = $repo;
}
/**
* Store a newly created resource in storage.
*
* #return Response
*/
public function store(Request $request)
{
$result = $this->repo->create($request);
return response()->json($result, 200);
}
}
class UserController {
// ... other code (including member variable $repo)
public function store(UserRequest $request)
{
$service = new StoringService($this->repo); // Or put this in your BaseController's constructor and make $service a member variable
return $service->store($request);
}
}
Using a trait
You can also use a trait, but you have to rename the trait's store() method then:
trait StoringTrait {
/**
* Store a newly created resource in storage.
*
* #return Response
*/
public function store(Request $request)
{
$result = $this->repo->create($request);
return response()->json($result, 200);
}
}
class UserController {
use {
StoringTrait::store as baseStore;
}
// ... other code (including member variable $repo)
public function store(UserRequest $request)
{
return $this->baseStore($request);
}
}
The advantage of this solution is that if you do not have to add extra functionality to the store() method, you can just use the trait without renaming and you do not have to write an extra store() method.
Using inheritance
In my opinion, inheritance is not so suitable for the kind of code reuse that you need here, at least not in PHP. But if you want to only use inheritance for this code reuse problem, give the store() method in your BaseController another name, make sure that all classes have their own store() method and call the method in the BaseController. Something like this:
BaseController.php
/**
* Store a newly created resource in storage.
*
* #return Response
*/
protected function createResource(Request $request)
{
$result = $this->repo->create($request);
return response()->json($result, 200);
}
UserController.php
public function store(UserFormRequest $request)
{
return $this->createResource($request);
}
You can move your logic from BaseController to trait, service, facade.
You can not override existing function and force it to use different type of argument, it would break stuff. For example, if you later would write this:
function foo(BaseController $baseController, Request $request) {
$baseController->store($request);
}
It would break with your UserController and OtherRequest because UserController expects UserController, not OtherRequest (which extends Request and is valid argument from foo() perspective).
As others have mentioned, you cannot do what you want to do for a host of reasons. As mentioned, you can solve this problem with traits or similar. I am presenting an alternative approach.
At a guess, it sounds like you are trying to follow the naming convention put forth by Laravel's RESTful Resource Controllers, which is kind of forcing you to use a particular method on a controller, in this case, store.
Looking at the source of ResourceRegistrar.php we can see that in the getResourceMethods method, Laravel does either a diff or intersect with the options array you pass in and against the default values. However, the those defaults are protected, and include store.
What this means is that you can't pass anything to Route::resource to force some override of the route names. So let's rule that out.
A simpler approach would be to simply set up a different method just for this route. This can be achieved by doing:
Route::post('user/save', 'UserController#save');
Route::resource('users', 'UserController');
Note: As per the documentation, the custom routes must come prior to the Route::resource call.
The declaration of UserController::store() should be compatible with BaseController::store(), which means (among other things) that the given parameters for both the BaseController as well as UserController should be exactly the same.
You actually cán force the BaseController to require a UserFormRequest, it's not the prettiest solution, but it works.
By overwriting there is no way you can replace Request with UserFormRequest, so why not use both? Giving both methods an optional parameter for injecting the UserFormRequest object. Which would result in:
BaseController.php
class BaseController {
public function store(Request $request, UserFormRequest $userFormRequest = null)
{
$result = $this->repo->create($request);
return response()->json($result, 200);
}
}
UserController.php
class UserController extends BaseController {
public function __construct(UserRepository $repo)
{
$this->repo = $repo;
}
public function store(UserFormRequest $request, UserFormRequest $userFormRequest = null)
{
return parent::store($request);
}
}
This way you can ignore the parameter when using BaseController::store() and inject it when using UserController::store().
The easiest and cleanest way I found to circumvent that problem was to prefix the parent methods with an underscore. For example:
BaseController:
_store(Request $request) { ... }
_update(Request $request) { ... }
UserController:
store(UserFormRequest $request) { return parent::_store($request); }
update(UserFormRequest $request) { return parent::_update($request); }
I feel like creating service providers is an overkill. What we're trying to circumvent here is not the Liskov substitution principle, but simply the lack of proper PHP reflection. Type-hinting methods is, in itself, a hack after all.
This will force you to manually implement a store and update in every child controller. I don't know if that's bothersome for your design, but in mine, I use custom requests for each controller, so I had to do it anyway.
I am trying to resolve class via __construct using Laravel's bind() method.
Here what I do:
routes.php (of course I will move it away from here)
// Bindings
App::bind(
'License\Services\ModuleSelector\SelectorInterface',
'License\Services\ModuleSelector\ModuleSelector'
);
SelectorInterface.php - interface that I will expect in __construct method.
<?php namespace License\Services\ModuleSelector;
interface SelectorInterface {
/**
* Simply return query that will select needle module fields
*
* #return mixed
*/
public function make();
}
ModuleSelector.php - this is class that I want to resolve via Laravel's DI (see example below).
<?php namespace License\Services\ModuleSelector;
use License\Services\ModuleSelector\Selector;
class ModuleSelector extends Selector
{
/**
* Get module by it's code
*
* #return mixed
*/
public function find()
{
return $this->make()
->where('code', $module_code)
->first();
}
}
Module.php
<?php namespace License\Services\ModuleType;
use License\Services\ModuleType\TypeInterface;
use License\Services\ModuleSelector\SelectorInterface;
class Module
{
...
function __construct(SelectorInterface $selector)
{
$this->selector = $selector;
}
...
}
And the place when error occurs:
In my repo I have use License\Services\ModuleType\Module as ModuleService;.
Than there is method called find():
/**
* Find module by its code with all data (types, selected type)
* #return mixed
*/
public function find($module_code)
{
$module = new ModuleService;
// Get module id in order to use build in relations in framework
$module = $this->module->find($module_code);
...
}
So, in other words, I have 2 classes and one interface. What I am trying to do is:
1) Create Class1.php / Class2.php / Class2Interface.php.
2) In Class1.php in the __construct I specify __construct(Class2Interface $class2).
3) Instantiate Class2.
What I am doing wrong? Examples found here.
In this line:
$module = new ModuleService;
You are directly invoking the Module class and not passing in an instance of SelectorInterface.
For the IoC to work you bind and make classes using it. Try that line again with :
$module = App::make('License\Services\ModuleSelector\SelectorInterface');
An alernative is to inject it directly into your repos constructor, as long as the repo is created by the IoC container, your concrete will be automatically injected.
Nowhere do you have a class marked to actually "implement SelectorInterface".
I want to extend/overwrite the method logAttempt in class Confide (Confide on GitHub) in order to execute some extra code whenever someone logs in successfully. This would be cleaner than copying the same code to all controllers where logAttempt is called.
I read through the Laravel documentation and several answers here on stackoverflow, but I just can't get it working.
I created a new folder app/extensions with a file named Confide.php:
<?php
namespace Extensions;
class Confide extends \Zizaco\Confide\Confide {
public function __construct(ConfideRepository $repo) {
die('no way!');
$this->repo = $repo;
$this->app = app();
}
public function logAttempt($credentials, $confirmed_only = false, $identity_columns = array()) {
die('yeah man!');
}
}
I added the directory to my app/start/global.php:
ClassLoader::addDirectories(array(
// ...
app_path().'/extensions',
));
I also added it to composer.json and ran composer dump-autoload:
"autoload": {
"classmap": [
...,
"app/extensions"
]
},
My own Confide class seems not to be loaded at all, because Confide works as normal – without ever die()-ing.
And if I use \Extensions\Confide::logAttempt($input, true); in my controller including the namespace, I get this ErrorException:
Non-static method Extensions\Confide::logAttempt() should not be called statically, assuming $this from incompatible context
Do I really need my own ConfideServiceProvider class as well? I tried that, too, but I'm not sure at all what to put in there to make Confide use my extended class.
Is there no simple way to extend a tiny bit of a class? There must be, I'm just missing something here.
If you are looking to execute some code when a user logs in, you should just listen for that event. In this case, I believe Confide uses the Auth class to login, so you should be able to listen for that event.
Event::listen('auth.login', function($user)
{
$user->last_login = new DateTime;
$user->save();
});
I find this much easier and cleaner than worrying about extending classes.
EDIT: Made a mistake
I think you need to call the method like this:
\Extensions\Confide->logAttempt($input, true);
because you are using:
\Extensions\Confide::logAttempt($input, true);
Which is how you call static methods.
I think I finally figured it out.
I had to extend ConfideServiceProvider as well like so:
<?php
namespace Extensions;
class ConfideServiceProvider extends \Zizaco\Confide\ConfideServiceProvider {
/**
* Bootstrap the service provider.
*
* #return void
*/
public function boot() {
$this->package('extensions/confide');
}
/**
* Register the application bindings.
*
* #return void
*/
protected function registerConfide() {
$this->app->bind('confide', function($app) {
return new Confide($app->make('confide.repository'));
});
}
}
The code above goes into app/extensions/ConfideServiceProvider.php. Note: In boot() I replaced "zizaco" with "extensions" and in registerConfide() I made no changes at all, but if this method is not present in the extended class, the original class will be used. I've got no idea why.
Then in app/config/app.php I replaced Zizaco\Confide\ConfideServiceProvider with Extensions\ConfideServiceProvider.
My own extended Confide class looks like this now:
<?php
namespace Extensions;
class Confide extends \Zizaco\Confide\Confide {
public function logAttempt($credentials, $confirmed_only = false, $identity_columns = array()) {
$result = parent::logAttempt($credentials, $confirmed_only, $identity_columns);
if ($result) {
// Login successful. Do some additional stuff.
\Log::info('User ' . \Auth::user()->username . ' logged in.');
}
return $result;
}
}
Note: If you want to use any other standard Laravel class like Log, Session etc., prefix it with one backslash as shown in the example above, or add a use operator for each class you use (e.g. use \Log;).
Commando need's help from you.
I have a controller in Yii:
class PageController extends Controller {
public function actionSOMETHING_MAGIC($pagename) {
// Commando will to rendering,etc from here
}
}
I need some magic method under Yii CController for controlling all subrequest under /page || Page controller.
Is this somehow possible with Yii?
Thanks!
Sure there is. The easiest way is to override the missingAction method.
Here is the default implementation:
public function missingAction($actionID)
{
throw new CHttpException(404,Yii::t('yii','The system is unable to find the requested action "{action}".',
array('{action}'=>$actionID==''?$this->defaultAction:$actionID)));
}
You could simply replace it with e.g.
public function missingAction($actionID)
{
echo 'You are trying to execute action: '.$actionID;
}
In the above, $actionID is what you refer to as $pageName.
A slightly more involved but also more powerful approach would be to override the createAction method instead. Here's the default implementation:
/**
* Creates the action instance based on the action name.
* The action can be either an inline action or an object.
* The latter is created by looking up the action map specified in {#link actions}.
* #param string $actionID ID of the action. If empty, the {#link defaultAction default action} will be used.
* #return CAction the action instance, null if the action does not exist.
* #see actions
*/
public function createAction($actionID)
{
if($actionID==='')
$actionID=$this->defaultAction;
if(method_exists($this,'action'.$actionID) && strcasecmp($actionID,'s')) // we have actions method
return new CInlineAction($this,$actionID);
else
{
$action=$this->createActionFromMap($this->actions(),$actionID,$actionID);
if($action!==null && !method_exists($action,'run'))
throw new CException(Yii::t('yii', 'Action class {class} must implement the "run" method.', array('{class}'=>get_class($action))));
return $action;
}
}
Here for example, you could do something as heavy-handed as
public function createAction($actionID)
{
return new CInlineAction($this, 'commonHandler');
}
public function commonHandler()
{
// This, and only this, will now be called for *all* pages
}
Or you could do something way more elaborate, according to your requirements.
You mean CController or Controller (last one is your extended class) ?
If you extended CController class like this:
class Controller extends CController {
public function beforeAction($pagename) {
//doSomeMagicBeforeEveryPageRequest();
}
}
you could get what you need