Php 无法使用 slim 的父级容器

我正在try 使用SLIM和下面的是我的家庭控制器来构建一个项目

HomeController.php个

<?php

namespace Controllers;

use DI\Container;
use Psr\Http\Message\ResponseInterface as Response;
use Psr\Http\Message\ServerRequestInterface as Request;
use Slim\Views\Twig;


class HomeController extends Controller{

   private $db;

   private $view;

   private $container;

    public function __construct(Container $container){
    
      $this->container = $container;

       $this->db = $container->get('db');

       $this->view = $container->get('view');
    }

    public function index(Request $request, Response $response, $args){

      echo $this->test; // 'hello world'; This is from Controller.php

      echo $this->container->get('container-test'); // I am from container is the string i pass is container and i am getting output.

      echo $this->container_from_controller?'Captured':'Not Captured'; // Not Captured i.e the container in Controller.php is empty
      
      return $this->view->render($response, 'home/index.html', [

          'name' => 'from home controller'

      ]);
    }
}

Controller.php个

<?php

namespace Controllers;

use DI\Container;



class Controller{

    protected $container_from_controller;

    protected $test = 'hello world';

    public function __construct(Container $container){
      
        $this->container_from_controller = $container;
    }


}

代码运行得很好.作为改进，我考虑将HomeController构造函数移到Controller.php，这样我就不想在每个控制器上重复构造代码.

也就是说我想搬家

   $this->container = $container;

   $this->db = $container->get('db');

   $this->view = $container->get('view');

到Controller.php，因此我可以直接访问$This-&>视图和$This-&>db，因为我可以访问$This-&>test

但我可以看到Controller.php__构造没有执行.

我try 将Parent::Construct()放入HomeController__Construct函数，但得到的参数太少错误.

访问控制器中的容器最好的方法是什么，所以我不想重复db和view代码.

composer.json个

{
    "require": {
        "slim/slim": "4.*",
        "slim/psr7": "^1.6",
        "php-di/slim-bridge": "^3.4",
        "dcblogdev/pdo-wrapper": "^2.0",
        "slim/twig-view": "^3.3"
    },
    "autoload": {
        "psr-4": {
            "Controllers\\": "src/Controllers"
        }
    }
}

Update of codes as per hakre answer个

我将控制器重命名为BaseController，这更有意义.

HomeController个

<?php

namespace Controllers;

use Psr\Http\Message\ResponseInterface as Response;
use Psr\Http\Message\ServerRequestInterface as Request;


class HomeController extends BaseController{



    public function index(Request $request, Response $response, $args){

      $sql = "SELECT * FROM list";

      $rows = $this->db->rows($sql);

      var_dump($rows);
      
      return $this->view->render($response, 'home/index.html', [

          'name' => 'from home controller'

      ]);
    }
}

BaseController个

<?php

namespace Controllers;

use DI\Container;



abstract class BaseController{

    protected $db;

    protected $view;

    public function __construct(Container $container){
      
        $this->db = $container->get('db');

        $this->view = $container->get('view');
    }


}