如何在Laravel 4中解决"目标[接口]不可实现"？

Question

我的错误信息:

Illuminate \ Container \ BindingResolutionException 
Target [Project\Backend\Service\Validation\ValidableInterface] is not instantiable.

我知道接口和抽象类是不可实例化的,所以我知道Laravel不应该尝试实例化我的接口.但不知何故,它正在努力,我怀疑这可能是一个具有约束力的问题......即使我相信我已经正确绑定它并将其注册为服务提供商.

我应该提一下,我从Chris Fidao的"实施Laravel"中取出了这个例子,它几乎完全相同!

这是我的表单类的前几行:

namespace Project\Backend\Service\Form\Job;

use Project\Backend\Service\Validation\ValidableInterface;
use Project\Backend\Repo\Job\JobInterface;

class JobForm {

    /**
     * Form Data
     *
     * @var array
     */
    protected $data;

    /**
     * Validator
     *
     * @var \Project\Backend\Form\Service\ValidableInterface
     */
    protected $validator;

    /**
     * Job repository
     *
     * @var \Project\Backend\Repo\Job\JobInterface
     */
    protected $job;

    public function __construct(ValidableInterface $validator, JobInterface $job)
    {
        $this->validator = $validator;
        $this->job = $job;
    }

这是我的验证器类的前几行:

namespace Project\Backend\Service\Form\Job;

use Project\Backend\Service\Validation\AbstractLaravelValidator;

class JobFormValidator extends AbstractLaravelValidator {

    // Includes some validation rules

这是抽象验证器:

namespace Project\Backend\Service\Validation;

use Illuminate\Validation\Factory;

abstract class AbstractLaravelValidator implements ValidableInterface {

    /**
     * Validator
     *
     * @var \Illuminate\Validation\Factory
     */
    protected $validator;

    /**
     * Validation data key => value array
     *
     * @var Array
     */
    protected $data = array();

    /**
     * Validation errors
     *
     * @var Array
     */
    protected $errors = array();

    /**
     * Validation rules
     *
     * @var Array
     */
    protected $rules = array();

    /**
     * Custom validation messages
     *
     * @var Array
     */
    protected $messages = array();

    public function __construct(Factory $validator)
    {
        $this->validator = $validator;
    }

这是我将其全部绑定到应用程序的代码:

namespace Project\Backend\Service\Validation;

use Illuminate\Support\ServiceProvider;
use Project\Backend\Service\Form\Job\JobFormValidator;

class ValidationServiceProvider extends ServiceProvider {

    public function register()
    {
        $app = $this->app;

        $app->bind('Project\Backend\Service\Form\Job\JobFormValidator', function($app)
        {
            return new JobFormValidator($app['validator']);
        });
    }
}

然后在app/config/app.php中注册:

.....
'Project\Backend\Service\Validation\ValidationServiceProvider',
....

最后这些是我的控制器的前几行:

use Project\Backend\Repo\Job\JobInterface;
use Project\Backend\Service\Form\Job\JobForm;

class JobController extends \BaseController {

    protected $jobform;

    function __construct(JobInterface $job, JobForm $jobform)
    {
        $this->job = $job;
        $this->jobform = $jobform;
    }

Answer 1

当通过类型提示将它注入构造函数时,您需要告诉Laravel它应该用于某个接口的实例.

您可以使用该bind()方法执行此操作(例如,在您的服务提供商中)

$app->bind('JobInterface', 'Job'); // Job being the class you want to be used

我强烈建议你观看视频,其中Laravel的创造者Taylor Otwell解释了这个以及其他一些事情.