Laravel 5.7：构建时目标不可实例化

Question

我知道有很多答案，但我无法真正解决这个问题。

我确实按照这个答案（How to make a REST API First Web application in Laravel）在 Laravel 5.7 上创建存储库/网关模式

我在 github 上还有“项目”，如果有人真的想要 test/clone/see ： https: //github.com/sineverba/domotic-panel/tree/development （开发分支）

应用\接口\Lan接口

<?php
/**
 * Interface for LAN models operation.
 */

namespace App\Interfaces;


interface LanInterface
{

    public function getAll();

}

应用\提供商\服务提供商

<?php

namespace App\Providers;

use Illuminate\Support\ServiceProvider;
use Illuminate\Support\Facades\Schema;

class AppServiceProvider extends ServiceProvider
{
    /**
     * Bootstrap any application services.
     *
     * @return void
     */
    public function boot()
    {
        /**
         * Solve the "Key too long" issue
         *
         * @see https://laravel-news.com/laravel-5-4-key-too-long-error
         */
        Schema::defaultStringLength(191);
    }

    /**
     * Register any application services.
     *
     * @return void
     */
    public function register()
    {
        $this->app->register(RepositoryServiceProvider::class);
    }
}

应用程序\提供者\RepositoryServiceProvider

<?php

namespace App\Providers;

use Illuminate\Support\ServiceProvider;

class RepositoryServiceProvider extends ServiceProvider
{

    public function register()
    {
        $this->app->bind(
            'app\Interfaces\LanInterface',           // Interface
            'app\Repositories\LanRepository'        // Eloquent
        );
    }

}

应用\网关\LanGateway

<?php

/**
 * The gateway talks with Repository
 */

namespace App\Gateways;
use App\Interfaces\LanInterface;


class LanGateway
{

    protected $lan_interface;

    public function __construct(LanInterface $lan_interface) {
        $this->lan_interface = $lan_interface;
    }

    public function getAll()
    {
        return $this->lan_interface->getAll();
    }

}

应用\存储库\LanRepository

<?php
/**
 * Repository for LAN object.
 * PRG paradigma, instead of "User"-like class Model
 */

namespace App\Repositories;
use App\Interfaces\LanInterface;
use Illuminate\Database\Eloquent\Model;


class LanRepository extends Model implements LanInterface
{

    protected $table = "lans";

    public function getAll()
    {
        return 'bla';
    }

}

App\Providers\RepositoryServiceProvider::class,我也在以下providers部分添加了config\app.php

这最终是控制器（我知道它还不完整）：

<?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;
use App\Gateways\LanGateway;

class LanController extends Controller
{

    private $lan_gateway;

    /**
     * Use the middleware
     *
     * @return void
     */
    public function __construct(LanGateway $lan_gateway)
    {
        $this->middleware('auth');
        $this->lan_gateway = $lan_gateway;
    }

    /**
     * Display a listing of the resource.
     *
     * @return \Illuminate\Contracts\Support\Renderable
     */
    public function index()
    {

        $this->lan_gateway->getAll();
        return view('v110.pages.lan');
    }
}

我得到的错误是

Target [App\Interfaces\LanInterface] is not instantiable while building [App\Http\Controllers\LanController, App\Gateways\LanGateway].

我确实尝试过：

php artisan config:clear php artisan clear-compiled

Answer 1

我认为@nakov 关于区分大小写的说法可能是正确的。我不相信 PHP 本身关心大小写命名空间，但 Composer 自动加载器和 Laravel 容器使用 key->value 数组键（它们确实具有区分大小写的键）来绑定和检索容器中的类。

为了确保名称始终匹配，请尝试使用特殊::class常量，如下所示：

<?php

namespace App\Providers;

use Illuminate\Support\ServiceProvider;
use App\Repositories\LanRepository;
use App\Interfaces\LanInterface;

class RepositoryServiceProvider extends ServiceProvider
{

    public function register()
    {
        $this->app->bind(
            LanInterface::class,
            LanRepository::class
        );
    }

}