Question

我正在尝试实施https://www.slimframework.com/docs/v3/objects/router.html所描述的Allow Slim to instantiate the controller部分。这样做时，我收到以下错误：

传递给Michael \ Test \ HomeController :: __ construct（）的参数1必须是Slim \ ContainerInterface的实例，Slim \ Container的实例给予，召唤 /var/www/slimtest/vendor/slim/slim/Slim/CallableResolver.php上线 93

认为它可能与命名空间有关，我也在\命名空间中尝试但是也遇到了同样的错误。

https://www.slimframework.com/docs/v3/objects/router.html上的文档是否不正确，HomeController的构造函数参数声明类型应该是Slim\Container，还是我做错了，Slim\ContainerInterface是否正确？

<?php
namespace Michael\Test;

error_reporting(E_ALL);
ini_set('display_startup_errors', 1);
ini_set('display_errors', 1);

require '../vendor/autoload.php';

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

$app = new \Slim\App();
$container = $app->getContainer();
//$container['view'] = function ($c) {};

//Question.  Do I need to use the fully qualified class name???
$app->get('/', \Michael\Test\HomeController::class . ':home');
//$app->get('/', '\Michael\Test\HomeController:home');

$app->run();

家庭控制器

namespace Michael\Test;
class HomeController
{
    protected $container;

    // constructor receives container instance
    public function __construct(\Slim\ContainerInterface $container) {
        $this->container = $container;
    }

    public function home($request, $response, $args) {
        // your code
        // to access items in the container... $this->container->get('');
        return $response;
    }

    public function contact($request, $response, $args) {
        // your code
        // to access items in the container... $this->container->get('');
        return $response;
    }
}

Answer 1

\Slim\ContainerInterface不存在（请参阅here）。查看\Slim\Container您需要使用的接口Interop\Container\ContainerInterface的实现，或者您可以使用Slim实现\Slim\Container作为类型参数。

Slim容器的正确接口是什么？

1 个答案: