I have a controller in my Laravel project called ImageController. This is a pretty basic CRUD controller.
When I access /images/{id} through my ImageController#show action, I want to also display comments. However, I don't want to put the comment logic in my ImageController. For this logic, I have created an ImageCommentController.
I'm not really sure how to go about this, but I'm trying to do something of this sort:
class ImageController extends BaseController {
// methods ...
public function show($id)
{
$images = // get images ...
$this->layout->view = // images.show and imagescomment.index (using ImageCommentsController#index logic)
}
}
I'm sorry if this is vaguely phrased, let me know if it is and I'll try to make it more understandable.
Maybe a better solutions than using a Controller for displaying the comments is to use a class with a method renderComments() that basically does something like:
class Comments {
public static renderComments($commentType = 'images')
{
$comments = Comments::where('comment_type', '=', $commentType)->get();
return View::make('comments', $comments)->render();
}
}
Then for example inside your image view:
...
{{ Comments::renderComments() }}
...
Related
Trying to load different views and methods depending on which view the user is browsing.
Views:
public function edit()
{
if("SOMETHING")return View::make('store_edit');
If("SOMETHING")return View::make('product_edit');
}
Methods:
public function destroy($id)
{
if(SOMETHING){
$store = Store::find($id);
$store->delete();
return redirect('/store');
}
if(SOMETHING){
$product = Product::find($id);
$product->delete();
return redirect('/product');
}
}
What can be used in the if() statements depending on which view is browsed in order to delete the right item and not having to rewrite the functions for each table.
There isn't a simple way to get information about which view was displayed in a previous request, and that's probably not what you want. You should create separate controllers/routes for both "products" and "store". Then you can do away with that view logic altogether.
To somewhat answer your question, you can access information about the current route with the Route facade.
$route = Route::current();
$name = Route::currentRouteName();
$action = Route::currentRouteAction();
Read Laravel routing:
https://laravel.com/docs/5.6/routing#route-parameters
Route
Route::get('something/{param}', 'SomeController#edit');
Controller
...
public function edit($param) {
if ($param === $expectedParam) {...} else {...}
}
Also you can make $param optional:
Route
Route::get('something/{param?}', 'SomeController#edit');
Controller (don't forget to give a default value)
...
public function edit($param = null) {
if ($param === $expectedParam) {...} else {...}
}
Please do not use same controller for diferent models.
if has diferent view and diferent functions this HAVE to has diferent controllers
Again, Laravel is a beautiful framework, dont do this!
I'm currently rebuilding my vanilla-PHP-App with Laravel and I have the following problem.
I have multiple database-tables, that represent word categories (noun, verb, adverb, ...). For each table I created a separate Model, a route::resource and a separate resource-Controller. For example:
NomenController.php
public function show($id)
{
$vocab = Nomen::find($id);
return view('glossarium.vocab_update', compact('vocab'));
}
and
VerbController.php
public function show($id)
{
$vocab = Verb::find($id);
return view('glossarium.vocab_update', compact('vocab'));
}
...which are essentially the same except the Model class.
I don't want to create a separate Controller for each model, that does exactly the same. What would be the most simple and elegant way to solve this?
Should I just create a VocabController.php and add a parameter for the Model-name like:
Route::resource('/vocab/{category}', 'VocabController');
and then add a constructor method in this controller like
public function __construct ($category) {
if ($category == 'nomen') {
$this->vocab = App\Nomen;
}
else if ($category == 'verb') {
$this->vocab = App\Verb;
}
}
I wonder if there is a simpler method to do that. Can I somehow do this with Route Model Binding?
Thanks in advance
Simply create a trait like this in App\Traits, (you can name it anything... Don't go with mine though... I feel its pretty lame... :P)
namespace App\Traits;
trait CommonControllerFunctions {
public function show($id) {
$modelObject = $this->model;
$model = $modelObject::find($id);
return view('glossarium.vocab_update', compact('model'));
}
}
and in your NomenController and VerbController, do this:
use App\Traits\CommonControllerFunctions;
class NomenController {
use CommonControllerFunctions;
protected $model = Nomen::class;
}
and
use App\Traits\CommonControllerFunctions;
class VerbController {
use CommonControllerFunctions;
protected $model = Verb::class;
}
Note: Please note that this example is just a work-around for your particular situation only... Everyone practices code differently, so this method might not be approved by all...
I think the simpliest way it to create only one controller, eg VocabController with methods nomen, verb and whatever you want.
Routes:
Route::get('/vocab/nomen/{nomen}', 'VocabController#item');
Route::get('/vocab/verb/{verb}', 'VocabController#item');
And the model binding:
Route::model('nomen', 'App\Nomen');
Route::model('verb', 'App\Varb');
Then your method shoud look like that:
public function item($item)
{
return view('glossarium.vocab_update', $item);
}
Keep in mind, that $item is already fetched model from the database.
I'm trying to load a blade view with layout, but I get this error:
"Attempt to assign property of non-object"
The structure is the following:
Route:
Route::pattern('controller', '\w+');
Route::get('{controller}', function($controller) {
$controllerClass = $controller.'Controller';
App::make($controllerClass)->index();
});
Controller:
class PricesController extends BaseController {
protected $layout = 'layouts.master';
public function index()
{
$this->layout->content = View::make('prices.index');
}
}
The debug says the issue is at line $this->layout->content = View::make('prices.index');
The views are fine... I have layouts folder with master.blade.php and I also have prices folder with index.blade.php.
The content section is exists as well with #stop and the #yield is there in the layout.
In the BaseController there is the setupLayout method:
protected function setupLayout()
{
if ( ! is_null($this->layout))
{
$this->layout = View::make($this->layout);
}
}
What is the problem? Why I get that exception?
Thank you!
I know I helped you in the #laravel irc channel but there are 3 things here for any others with this problem.
This is not a good use of route files. Controller implicit routing is hard to maintain if your app gets larger. Consider using Route::resource instead if you're just trying to save a few lines of code. But I'll give you the benefit of the doubt.
You'll want to use the nest method for your layout, i.e. $this->layout->nest('content', 'prices.index');
The setupLayout() function is not being called because you are calling index() directly on the object. This is not how Laravel normally processes controllers.
I'm not going to walk through the entire routing process but if you look at vendors/laravel/framework/src/Illuminate/Routing/ControllerDisplatcher.php on line 89 you will see:
protected function call($instance, $route, $method)
{
$parameters = $route->parametersWithoutNulls();
return $instance->callAction($method, $parameters);
}
Let's look at vendors/laravel/framework/src/Illuminate/Routing/Controller.php on line 227 and you will see:
public function callAction($method, $parameters)
{
$this->setupLayout();
$response = call_user_func_array(array($this, $method), $parameters);
... irrelevant stuff omitted ...
}
These reason I show these things is show the magic Laravel is doing behind the scenes.
Basically you are skipping that magic and just calling PricesController->index() directly instead of going through the router. This is why setupLayout is never being called and you will get an exception because there is no $this->layout object yet created.
I like MVC (a lot), and I am trying to teach myself a framework of MVC architecture in all the major web languages of today.
I am currently on CodeIgniter and PHP. I searched online for a way to make the same function behave different for a POST and GET but couldn't find anything. Does CodeIgniter have this feature?
If you've used Ruby On Rails or ASP.NET MVC you'll know what I'm talking about, in them frameworks we can do this:
[GET]
public ActionResult Edit(int Id)
{
// logic here for GET
}
[POST]
public ActionResult Edit(EntityX EX)
{
// logic here for POST
}
I am so used to this, that I am finding it hard wrapping my head around how to get the same smooth functionality without that useful ability.
Am I missing something? How can I achieve the same thing in CodeIgniter?
Thanks
Am I missing something? How can I achieve the same thing in
CodeIgniter?
if you want to learn how to truly approach MVC in PHP, you can learn it from Tom Butler articles
CodeIgniter implements Model-View-Presenter pattern, not MVC (even if it says so). If you want to implement a truly MVC-like application, you're on the wrong track.
In MVP:
View can be a class or a html template. View should never be aware of a Model.
View should never contain business logic
A Presenter is just a glue between a View and the Model. Its also responsible for generating output.
Note: A model should never be singular class. Its a number of classes. I'll call it as "Model" just for demonstration.
So it looks like as:
class Presenter
{
public function __construct(Model $model, View $view)
{
$this->model = $model;
$this->view = $view;
}
public function indexAction()
{
$data = $this->model->fetchSomeData();
$this->view->setSomeData($data);
echo $this->view->render();
}
}
In MVC:
Views are not HTML templates, but classes which are responsible for presentation logic
A View has direct access to a Model
A Controller should not generate a response, but change model variables (i.e assign vars from $_GET or $_POST
A controller should not be aware of a view
For example,
class View
{
public function __construct(Model $model)
{
$this->model = $model;
}
public function render()
{
ob_start();
$vars = $this->model->fetchSomeStuff();
extract($vars);
require('/template.phtml');
return ob_get_clean();
}
}
class Controller
{
public function __construct(Model $model)
{
$this->model = $model;
}
public function indexAction()
{
$this->model->setVars($_POST); // or something like that
}
}
$model = new Model();
$view = new View($model);
$controller = new Controller($model);
$controller->indexAction();
echo $view->render();
The parameters only allow you to retrieve GET variables. If you want to get the POST variables, you need to use the Input library which is automatically loaded by CodeIgniter:
$this->input->post('data');
So, in your case, it would be:
public function edit($id = -1)
{
if($id >= 0 && is_numeric($id))
{
// logic here for GET using $id
}
else if($id === -1 && $this->input->post('id') !== false)
{
// logic here for POST using $this->input->post('id')
}
}
Note that you can also use this library to obtain GET, COOKIE and SERVER variables:
$this->input->get('data');
$this->input->server('data');
$this->input->cookie('data');
I have simple function in one of my controller (members):
function action_data($array) {
$datamembre = DB::table('members')->where('id', '=', $id)->first();
return $datamembre;
}
I want use it in the view of another controller, I do this in my template:
$datamembers = Controller::call('members#data', array($members_id));
Is there any better or proper way to do it?
The best would be to put it into a Member model IMO.
Use view composer http://laravel.com/docs/views#view-composers
View::composer(array('home', 'profile'), function($view)
{
//
});
update
You can put the method in BaseController, all controllers extend it, so you could get this method in any controller/view. And make it static.
// in BaseController.php (L4) or Base.php (L3)
static public function action_data($array) {
$datamembre = DB::table('members')->where('id', '=', $id)->first();
return $datamembre;
}
In the view
BaseController::action_data($array); //L4
Base_Cotroller::action_data($array); //L3
Hope it works. Haven't tested it.