Question

我知道答案很多，但是我真的无法解决。

我确实按照此答案（How to make a REST API first web application in Laravel）在Laravel 5.7上创建了存储库/网关模式

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

App \ Interfaces \ LanInterface

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

namespace App\Interfaces;


interface LanInterface
{

    public function getAll();

}

App \ Providers \ ServiceProvider

<?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);
    }
}

App \ Providers \ 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
        );
    }

}

App \ Gateways \ 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();
    }

}

App \ Repositories \ 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';
    }

}

我确实在App\Providers\RepositoryServiceProvider::class,的{{1}}部分中添加了providers

这终于是控制器了（我知道它还不完整）：

config\app.php

我得到的错误是

<?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');
    }
}

我确实尝试过：

Target [App\Interfaces\LanInterface] is not instantiable while building [App\Http\Controllers\LanController, App\Gateways\LanGateway]. php artisan config:clear

Answer 1

我认为@nakov区分大小写可能是正确的。我认为PHP名称空间不区分大小写，但是当由作曲家自动加载或从Laravel容器解析时，它们是（因为匹配的数组键是一个字符串）。

尝试一下：

<?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
        );
    }

}

Answer 2

在我的情况下，我忘记将提供者登记到 confit/app.php 这就是错误的原因。

Laravel 5.7：在构建时无法实例化目标

2 个答案: