CODEWITHSUNDEEP
php
PHPDEVLOPER
PHPDEVLOPER

Reputation: 161

Rendering a view file in self made PHP MVC framework

I've developed My Own MVC Framework using php. I call view files in controller like:

include('../view/home.php');

but I want to use it like:

$this->view('home');

How can I define common function for that where I can just pass view name i.e home only and it will do include view file without passing the full file path?

Upvotes: 2

Views: 1420

Answers (2)

Peshraw H. Ahmed
Peshraw H. Ahmed

Reputation: 469

No one could answer you without seeing your codes really. But this should be my approach. You should have a class that all your controllers extend. Lets say that you have class Controllers and all your controllers extend it.

Then you may have a method inside the class named view($view_name).

public function view($view_name){
   include $some_path . '/' . $view_name . '.php';
}

then whenever you call view by $this->view it will include the view if it exists. This is not the best approach and I did not test the code. I just wanted to show you the path

Upvotes: 3

num8er
num8er

Reputation: 19372

I don't know Your MVC file/directory/namespace and etc structure.

But for beginner who tries to learn MVC and Frameworks by "reinventing wheel" (: I can give such example:

1) Create common abstract controller class in app/controllers folder:

namespace App\Controllers;

abstract class Controller {

  public function view($name) {
    include(__DIR__.'/../views/'.$name.'.php';
  }

}

2) Create Your own controller i.e. PagesController and use it:

namespace App\Controllers;

class PagesController extends Controller {

  public function home() {
    $this->view('home');
  }
}

p.s. You may omit namespace-ing, abstract word depending on autoloader logic

Upvotes: 2

Related Questions