laravel 将查询功能从路由移动到存储库

Question

嗨，我对 laravel 的世界还很陌生，我正在建立一个存储库，因为我正在重复对数据库的如此多的调用，所以将它们全部保存在一个地方并参考这个是有意义的。无论如何，我有一个链式选择，它查看 client_id 并找到与该客户端匹配的项目。我让它在我的 routes.php 文件中工作：

 Route::get('task/clientsprojects', function(){
       $input = Input::get('option');
       $client = Client::find($input);
       $projects =  $projects = Project::where('client_id', $client->id) 
                        ->orderBy('project_name') 
                        ->get(array('id','project_name') 

                        );
   $response = array(); 
   foreach($projects as $project){ 
   $response[$project->id] = $project->project_name; 
  } 

  return Response::json($response);
});

我的 create.blade.php 文件中有一个：

   <!--These two select boxes are linked together --->
@if(count($client_options)>0)
        {{ Form::select('client', $client_options, Input::old('client'), array('data-placeholder' => 'Choose a Client...', 'id' => 'select_client', 'class' => 'chosen-select tempus_select', 'tabindex' => '2', )) }}
    @endif 

 {{ Form::select('project', array_merge(array('default' => 'Select client first')), 'default', array('class' => 'tempus_select', 'id' => 'project_select')) }}

<script>
$(document).ready(function($){ 
    $('#select_client').change(function(){ 
        $.get("/task/clientsprojects",{ 
            option: $(this).val() 
            }, function(data) { 
            console.log(data); 
            var model = $('#project_select'); 
            model.empty(); 
            $.each(data, function(key, value) { 
$('#project_select').append("<option value='"+key+"'>"+value+"</option>'");
            });
            $("#project_select").trigger("change");
        }); 
    });
});

</script>

我在我的存储库类中创建了这个函数：

 //fetch clients
public function getClients() 
    {

        return \Auth::user()->clients()->orderBy('client_name', 'asc')->lists('client_name','id');;
    }

 //fetch clients projects
public function getClientsProjects() {

    $input = Input::get('option');
    $client = Client::find($input);

    $projects = $projects = \Auth::user()->projects()->where('client_id', $client->id)->orderBy('project_name')->get(array('id','project_name'));

        $response = array(); 

        foreach($projects as $project){ 
        $response[$project->id] = $project->project_name; 
        } 

        return Response::json($response);
    }

我的控制器是这样引用 repo 的：

<?php
use Acme\Repositories\ProjectRepositoryInterface;
class TaskController extends \BaseController {

    public function __construct(ProjectRepositoryInterface $project) {

        $this->project = $project;

    }

public function create()
{
    //
     $tasks = Auth::user()->tasks;  

    $client_options = $this->project->getClients();


    return View::make('tasks.create', array( 'client_options' => $client_options, 'status' => $status, 'priority' => $priority));
}       

}

我现在如何将此选择路由到此函数以通过 ajax 检索数据？有谁知道我该如何进行？在上面的例子中，我在任务控制器中，我在 create 函数中获取我的客户端，在当前的实现中，我有 routes.php 捕获这个并执行查询我想将它更改为 repo，但我不确定如何实现这一点。

更新

我将我的 repo 更新为以下内容：

public function getClientsProjects() {

    $input = Input::get('option'); //line 42
    $client = Client::find($input);

$projects =  $projects = Project::where('client_id', $client->id) 
                           ->orderBy('project_name') 
                          ->get(array('id','project_name'));
        $response = array(); 

        foreach($projects as $project){ 
        $response[$project->id] = $project->project_name; 
        } 

        return $response;    
}

并将以下函数插入到我的控制器中：

public function clientsProjects()
{
    return Response::json($this->project->getClientsProjects());
} 

这是我的 routes.php 文件：

Route::get('task/clientsprojects', 'TaskController@clientsProjects');

但我在控制台中收到此错误：

Failed to load resource: the server responded with a status of 500 (Internal Server Error)     
http://tempus.local/task/clientsprojects?option=1
XHR finished loading: GET "http://tempus.local/task/clientsprojects?option=1".
Failed to load resource: the server responded with a status of 500 (Internal Server Error) 

Laravel 日志：

[2014-09-03 20:20:45] production.ERROR: exception 
'Symfony\Component\Debug\Exception\FatalErrorException' with message 'Class 
'Acme\Repositories\Input' not found' in 
/media/sf_Sites/tempus/app/Acme/Repositories/DbProjectRepository.php:42
 Stack trace:
#0 [internal function]: Illuminate\Exception\Handler->handleShutdown()
 #1 {main} [] []

Answer 1

您必须首先从存储库中仅返回相关数据：

public function getClientsProjects($clientId) 
{
    $client = Client::find($clientId);

    ...

    return $response;
}

然后改变你的路由指向你的控制器：

Route::get('task/clientsprojects', 'TaskController@clientsProjects');

然后在您的控制器中执行以下操作：

use Input; /// before your class declaration

...

public function clientsProjects()
{
    $clientId = Input::get('option');

    return Response::json($this->project->getClientsProjects($clientId));
}