在Laravel 5中找不到用户类

我遇到以下问题。我刚刚从4.2升级到Laravel 5，但我刚复制到新的Controllers文件夹的UserController不起作用。

它总是告诉我，它不会破坏用户模型。

当我复制我的代码时，它看起来如下：

UserController.php

<?php
use App'Transformer'UserTransformer;
class UserController extends ApiController
{
public function index()
{
    App::abort(403, 'This is not allowed');
}
public function show($id)
{   
    $user = User::find($id);
    if($user->access_token!=Input::get('access_token')){
        return "It is not allowed to access data from other users";
    }else{
        $result = $this->respondWithItem($user, new UserTransformer);
        return $result;
    }
}
public function getID()
{   
    $user = User::where('email', '=', Input::get('email'))->first();
    if($user->access_token!=Input::get('access_token')){
        return "It is not allowed to access data from other users";
    }else{
        $result = $user->id;
        return $result;
    }
}
public function register()
{
    $user = new User;
    $name = Input::get('name');
    $email = Input::get('email');
    $password = Input::get('password');
    if($name != null AND $email != null AND $password != null){
        if(User::where('email', '=', Input::get('email'))->exists()){
            return "A user with this email already exists";
        }else{
            $user->name = $name;
            $user->email = $email;
            $user->password = $password;
            $user->save();
            return "Success";
        }
    }else{
        return "All fields are required";
    }
}
public function login(){
    $user = User::where('email', '=', Input::get('email'))->first();
    if($user != null){
        if($user->password==Input::get('password')){
            $then =$user->time;
            $now = strtotime("now");
            $thenTimestamp = strtotime($then);
            $difference = $now - $thenTimestamp;
            if((empty($user->date) AND empty($user->time))OR($difference >=10800)OR($user->date!=date('d.m.Y'))){
                $user->access_token=self::getGUID();
                $timestamp = time();
                $user->date=date("d.m.Y",$timestamp);
                $timestamp = strtotime("now");
                $user->time=date('H:i', $timestamp);
                $user->save();
            }
            return $user->access_token;
        }
        else
            return "Wrong username or password";
    }else
        return "Wrong username or password";
}
function getGUID(){
    if (function_exists('com_create_guid')){
        return com_create_guid();
    }else{
        mt_srand((double)microtime()*100000);
        $charid = strtoupper(md5(uniqid(rand(), true)));
        $hyphen = chr(45);
        $uuid = 
            substr($charid, 0, 8).$hyphen
            .substr($charid, 8, 4).$hyphen
            .substr($charid,12, 4).$hyphen
            .substr($charid,16, 4).$hyphen
            .substr($charid,20,12);
        return $uuid;
    }
}
public function check(){
    $user = User::where('access_token', '=', Input::get('access_token'))->first();
    if($user != null){
    }else
    return "no session";
}
}

我并没有为控制器命名，用户模型只是开箱即用。UserTransformer中确实包含用户模型。这里是原件：

<?php
use App'User;
use League'Fractal'TransformerAbstract;
class UserTransformer extends TransformerAbstract
{
/**
 * Turn this item object into a generic array
 *
 * @return array
 */
public function transform(User $user)
{
    return [
        'id'             => (int) $user->id,
        'name'           => $user->name,
        'email'            => $user->email,
        'phone'         => $user->img,
        'mobile'       => $user->mobile,
        'address'       => $user->address,
    ];
}
}

这就是我工作所需要的一切。

但现在在Laravel 5中，UserController确实告诉我找不到User类。我尝试在UserController中添加类似"使用App/User"或"使用App/Http/Controllers/User"的行。它不起作用。我不知道为什么，因为我已经包括了我的变压器，我没有得到任何错误，但用户不工作。我也尝试了不同的使用方法来将用户模型包含到转换器中，因为我认为可能会有错误，但我没有得到一个工作版本。

我希望有人能帮助我。