Php “我该如何解决?”;目标[接口]不可实例化“;在拉威尔4号?
我的错误消息:Php “我该如何解决?”;目标[接口]不可实例化“;在拉威尔4号?,php,laravel,laravel-4,inversion-of-control,ioc-container,Php,Laravel,Laravel 4,Inversion Of Control,Ioc Container,我的错误消息: 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不应该试图实例化我的接口。然而,不知何故,它试图,我怀疑这可能是一个约束问题…尽管我相信我已经正确地约束了它,并已将其注册为服务提供商 我应该提到的是
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不应该试图实例化我的接口。然而,不知何故,它试图,我怀疑这可能是一个约束问题…尽管我相信我已经正确地约束了它,并已将其注册为服务提供商
我应该提到的是,我从Chris Fidao的“实现Laravel”中选取了这个例子,它几乎是一样的
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validator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;
}
这是我将所有内容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;
}
当通过类型暗示将某个接口注入构造函数时,您需要告诉Laravel它应该为该接口使用哪个实例 您可以使用
bind()
方法执行此操作(例如,在您的服务提供商中)
我强烈建议您观看Laravel的创建者Taylor Otwell解释这一点和其他一些事情的地方。首先,您需要使用 /app/Providers/AppServiceProvider.php
<?php namespace App\Providers;
use Illuminate\Support\ServiceProvider;
class AppServiceProvider extends ServiceProvider {
/**
* Bootstrap any application services.
*
* @return void
*/
public function boot()
{
//
}
/**
* Register any application services.
*
* @return void
*/
public function register()
{
//
$this->app->bind('JobInterface', 'Job');
}
}
有用的视频,并解释关键概念。但是我仍然不清楚,因为ValidableInterface仅由抽象类实现。然后,在将其添加到bindings.php之后,指定扩展抽象类的类(并因此实现接口),然后需要运行composer update命令
<?php namespace App\Providers;
use Illuminate\Support\ServiceProvider;
class AppServiceProvider extends ServiceProvider {
/**
* Bootstrap any application services.
*
* @return void
*/
public function boot()
{
//
}
/**
* Register any application services.
*
* @return void
*/
public function register()
{
//
$this->app->bind('JobInterface', 'Job');
}
}