Best practices to define response in action controller Laravel - laravel

I have a method inside PostController
class PostController extends Controller {
public function index() {
$posts = Post::all();
return response($posts);
}
}
Two way:
class PostController extends Controller {
public function index() {
$posts = Post::all();
return $posts;
}
}
Both work fine but which way is better and more correctly?

I personnaly prefer this version:
return \Response::json($data);
Because it makes clear that the response is actual json data.
Just make sure your code is understandable by someone new on your project.
If you are writing an API project, where everything is always returned in json, simply return the model because you don't have to make the reader learn that it's JSON because everything is in JSON everywhere.
On the other hand, if it's some sort of mixed project (some routes return view, JSON, XML, whatever), try to make is as obvious as you can that this specific route return JSON data and nothing else.
Also as stated in comments, stay consistent. The shorter isn't the better. The better is the one that is simple to read and give enough info about what's going on.

It doesn't matter actually.. its depends on your desire and consistency..
return Response::json($model);
//or
return response()->json($model);
//or
return $model;
it give you same output..but if you using response, you can set the status code. most of the time, this reponse thing is used in API project.

Related

Can a Laravel POST route return a view instead of redirecting to another route?

I have seen/worked on lot of projects where a Laravel POST route looking like this:
Route::post('/some-url', [SomeController::class, 'someMethod']);
And the controller with its method which takes care of the route looking like this:
class SomeController extends Controller
{
public function someMethod(Request $request) {
// My Logic to do something with the post data - $request
return redirect("/some-other-url");
}
}
have always redirected to a specific URL after doing something with the POST data, by convention.
I would like to know, if after processing a POST request, if it's okay to just return a view like what we do with GET requests? I know it works, but is it just convention or are there any issues in doing so?
Eg:
class SomeController extends Controller
{
public function someMethod(Request $request) {
// My Logic to do something with the post data - $request
return view("somebladeview")->with(["result" => $result ]);
}
}
The pattern is called post-redirect-get, or PRG. It is convention because POST should be used for requests that change data, whereas GET should be used to simply display data.
If you return a view after a POST, the user can hit reload, and (probably unintentionally) change data again, eg buy something a second time. By using PRG, the user hitting reload simply reloads the last GET, which just re-displays something.
NOTE: There is some advice in the comments on your question to ignore this convention. I think that is unwise and dangerous, these conventions exist for good reasons.

Returning same variable to every controller in laravel

I need to send the same result to almost every view page, so I need to bind the variables and return with every controller.
My sample code
public function index()
{
$drcategory = DoctorCategory::orderBy('speciality', 'asc')->get();
$locations = Location::get();
return view('visitor.index', compact('drcategory','locations'));
}
public function contact()
{
$drcategory = DoctorCategory::orderBy('speciality', 'asc')->get();
$locations = Location::get();
return view('visitor.contact', compact('drcategory','locations'));
}
But as you see, I need to write same code over and over again. How can I write it once and include it any function whenever I need?
I thought about using a constructor, but I cannot figure out how I can implement this.
You are able to achieve this by using the View::share() function within the AppServicerProvider:
App\Providers\AppServiceProvider.php:
public function __construct()
{
use View::Share('variableName', $variableValue );
}
Then, within your controller, you call your view as normal:
public function myTestAction()
{
return view('view.name.here');
}
Now you can call your variable within the view:
<p>{{ variableName }}</p>
You can read more in the docs.
There are a few ways to implement this.
You can go with a service, a provider or, like you said, within the constructor.
I am guessing you will share this between more parts of your code, not just this controller and for such, I would do a service with static calls if the code is that short and focused.
If you are absolutely sure it is only a special case for this controller then you can do:
class YourController
{
protected $drcategory;
public function __construct()
{
$this->drcategory = DoctorCategory::orderBy('speciality', 'asc')->get();
}
// Your other functions here
}
In the end, I would still put your query under a Service or Provider and pass that to the controller instead of having it directly there. Maybe something extra to explore? :)
For this, you can use View Composer Binding feature of laravel
add this is in boot function of AppServiceProvider
View::composer('*', function ($view) {
$view->with('drcategory', DoctorCategory::orderBy('speciality', 'asc')->get());
$view->with('locations', Location::get());
}); //please import class...
when you visit on every page you can access drcategory and location object every time
and no need to send drcategory and location form every controller to view.
Edit your controller method
public function index()
{
return view('visitor.index');
}
#Sunil mentioned way View Composer Binding is the best way to achieve this.

Laravel Backpack - getting current record from crud controller

In my crud controller I am trying to get the name of the person who is currently being edited.
so
http://192.168.10.10/admin/people/93/edit
In the people crud controller
public function setup() {
dd(\App\Models\People::get()->first()->name)
}
This returns the first person not the person currently being edited.
How do I return the current person (with an id of 93 in this example)
Ok, So since you use backpack look into CrudController to see how the method looks:
public function edit($id)
{
$this->crud->hasAccessOrFail('update');
$this->data['entry'] = $this->crud->getEntry($id);
$this->data['crud'] = $this->crud;
$this->data['fields'] = $this->crud->getUpdateFields($id);
$this->data['id'] = $id;
return view('crud::edit', $this->data);
}
So now you can overwrite the edit function and change whatever you want. You can even create a custom edit page if you so wish.
Setup on the other hand is usually used to add things like
$this->crud->addClause(...);
Or you can even get the entire constructor and put it in the setup method because setup call looks like this:
public function __construct()
{
// call the setup function inside this closure to also have the request there
// this way, developers can use things stored in session (auth variables, etc)
$this->middleware(function ($request, $next) {
$this->setup();
return $next($request);
});
}
So you could do something like \Auth::user()->id;
Also it's normal to work like this. If you only use pure laravel you will only have access to the current id in the routes that you set accordingly.
Rahman said about find($id) method. If you want to abort 404 exception just use method findOrFail($id). In my opinion it's better way, because find($id)->name can throw
"Trying to get property of non-object error ..."
findOrFail($id) first fetch user with specified ID. If doesn't exists just throw 404, not 500.
The best answer is:
public function edit($id)
{
return \App\Models\People::findOrFail($id);
}
Good luck.
you need person against id, try below
public function setup($id) {
dd(\App\Models\People::find($id)->name);
}

Is there any decent way to Decorate models returned from a Magento `[model]_load_after`event?

I'm trying to overwrite some methods in models, and I'm on a mission to avoid overwrites and rewrites of models for maximum compatibility with other modules.
I figured the best way would be to simply decorate models after they are loaded from Magento, however as far as I can tell because of the way the observer pattern in Magento is written it's impossible to accomplish this. ( As Magento always returns the reference to $this ), and the lack of interfaces might also cause trouble later down the road? See this partial of Mage/Core/Model/Abstract.php
/**
* Processing object after load data
*
* #return Mage_Core_Model_Abstract
*/
protected function _afterLoad()
{
Mage::dispatchEvent('model_load_after', array('object'=>$this));
Mage::dispatchEvent($this->_eventPrefix.'_load_after', $this->_getEventData());
return $this;
}
My question boils down to the title, is there a decent way of accomplishing this?, or am I simply stuck with rewrites :(?
The path I would like to take is;
On event [model]_load_after
return new Decorator($event->getObject())
Where the decorator class in my case would be something like;
public function __construct(Mage_Sales_Model_Order_Invoice $model)
{
parent::__construct($model); // sets $this->model on parent class, see below
}
// overwrite the getIncrementId method
public function getIncrementId()
{
return '12345';
}
// partial of parent class
public function __call($method, array $args)
{
return call_user_func_array(array($this->model, $method), $args);
}
And just some pseudo-code for extra clarification;
$model = Mage::getModel('sales/order_invoice')->load(1);
echo get_class($model);
Namespace_Decorator **INSTEAD OF** Mage_Sales_Model_...
echo $model->getIncrementId();
'12345' **INSTEAD OF** '1000001' ( or whatever the format might be )
Thanks for your time reading / commenting, I really hope there actually is a way to accomplish this in a clean fashion without making use of code overrides or rewrites of models.
Edit: extra clarification
Basically what I would like is to return an instance of the Decorator in a few cases, the sales_invoice being one of them and customer the other. So when any load() call is made on these models, it will always return the instance of the Decorator instead of the Model. Only method calls that the decorator overrides would be returned, and any other method calls would "proxied" through __call to the decorated object.
I'm not sure if I got your question right but here goes.
I think you can use the event [model]_load_after and simply do this:
$object = $event->getObject();
$object->setIncrementId('12345');
Or if you want to use a decorator class make it look like this:
public function __construct(Mage_Sales_Model_Order_Invoice $model)
{
parent::__construct($model);
$model->setIncrementId($this->getIncrementId());
}
public function getIncrementId()
{
return '12345';
}
I know that this is not exactly a decorator pattern but it should work.
I know that when adding a new method to the 'decorator' class you need to add it to attach data to the main model.
This is just my idea. I haven't got an other.
[EDIT]
You can try to rewrite the load method on the object to make it return what you need. But I wouldn't go that way. You can end up screwing a lot of other things.
I don't think there is an other way to do it because load always returns the current object no mater what you do in the events dispatched in the method. see Mage_Core_Model_Abstract::load()
public function load($id, $field=null)
{
$this->_beforeLoad($id, $field);
$this->_getResource()->load($this, $id, $field);
$this->_afterLoad();
$this->setOrigData();
$this->_hasDataChanges = false;
return $this;
}
By making it return new Decorator($this), you might achieve what you need, but just make sure that when calling $model->doSomething() and doSomething() is not a method in your decorator you still end up calling the original method on the model.

Should I call redirect() from within my Controller or Model in an MVC framework?

I'm using the MVC PHP framework Codeigniter and I have a straight forward question about where to call redirect() from: Controller or Model?
Scenario:
A user navigates to www.example.com/item/555. In my Model I search the item database for an item with the ID of 555. If I find the item, I'll return the result to my controller. However, if an item is not found, I want to redirect the user somewhere. Should this call to redirect() come from inside the model or the controller? Why?
No your model should return false and you should check in your controller like so:
class SampleModel extends Model
{
//Construct
public function FetchItem($id)
{
$result = $this->db->select("*")->from("table")->where("item_id",$id)->get();
if($result->num_rows() == 0)
{
return false;
}
//return result
}
}
and within your controller do:
function item($id)
{
$Item = $this->SampleModel->FetchItem($id);
if(!$Item)
{
redirect("class/error/no_item");
}
}
Models are for data only either return a standard result such as an key/value object or a boolean.
all logic should be handled / controlled by the Controller.
Models are not page specific, and are used globally throughout the whole application, so if another class / method uses the model, it might get redirect to the incorrect location as its a different part of your site.
It seems like the controller would be the best place to invoke your redirect because the controller typically delegates calls to the model, view, or in your case, another controller.
However, you should use whatever makes the most sense for your application and for what will be easier to maintain in the future, but also consider that rules do exist for a reason.
In short, if a coworker were to try to fix a bug in your code, what would the "reasonable person" standard say? Where would most of them be most likely to look for your redirect?
Plus, you said you're returning the result to your controller already... perhaps that's where you should make your redirect...

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