找不到类“角色” [Symfony \\ Component \\ Debug \\ Exception \\ FatalErrorException]

Question

因此，我已经看到很多关于类未找到错误的类似问题，但是除非我完全缺少明显的东西，否则我将无法理解以下方法调用如何看不到我的角色类以及导致该类未找到的结果例外：

 $user->makeEmployee("admin")

这是我的带有makeEmployee（）的用户类：

<?php namespace App;
use Illuminate\Auth\Authenticatable;
use Illuminate\Database\Eloquent\Model;
use Illuminate\Auth\Passwords\CanResetPassword;
use Illuminate\Contracts\Auth\Authenticatable as AuthenticatableContract;
use Illuminate\Contracts\Auth\CanResetPassword as CanResetPasswordContract;

class User extends Model implements AuthenticatableContract, CanResetPasswordContract {

use Authenticatable, CanResetPassword;

/**
 * The database table used by the model.
 *
 * @var string
 */
protected $table = 'users';

/**
 * The attributes that are mass assignable.
 *
 * @var array
 */
protected $fillable = ['name', 'email', 'password'];

/**
 * The attributes excluded from the model's JSON form.
 *
 * @var array
 */
protected $hidden = ['password', 'remember_token'];


/**
 * Get the roles a user has
 */
public function roles()
{
    return $this->belongsToMany('Role', 'users_roles');
}

/**
 * Find out if User is an employee, based on if has any roles
 *
 * @return boolean
 */
public function isEmployee()
{
    $roles = $this->roles->toArray();
    return !empty($roles);
}

/**
 * Find out if user has a specific role
 *
 * $return boolean
 */
public function hasRole($check)
{
    return in_array($check, array_fetch($this->roles->toArray(), 'name'));
}

/**
 * Get key in array with corresponding value
 *
 * @return int
 */
private function getIdInArray($array, $term)
{
    foreach ($array as $key => $value) {
        if ($value == $term) {
            return $key;
        }
    }

    throw new UnexpectedValueException;
}

/**
 * Add roles to user to make them a concierge
 */
public function makeEmployee($title)
{
    $assigned_roles = array();

    $roles = array_fetch(Role::all()->toArray(), 'name');

    switch ($title) {
        case 'admin':
            $assigned_roles[] = $this->getIdInArray($roles, 'create_message');
        /*case 'member':
            $assigned_roles[] = $this->getIdInArray($roles, 'create_customer');
        case 'concierge':
            $assigned_roles[] = $this->getIdInArray($roles, 'add_points');
            $assigned_roles[] = $this->getIdInArray($roles, 'redeem_points');*/
            break;
        default:
            throw new \Exception("The employee status entered does not exist");
    }

    $this->roles()->attach($assigned_roles);
}
}

这是我的角色类：

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class Role extends Model
{
    protected $table = 'roles';
    /**
     * Set timestamps off
     */
    public $timestamps = false;

    /**
     * Get users with a certain role
     */
    public function users()
    {
        return $this->belongsToMany('User', 'users_roles');
    }
}

如果它们都是相同的名称空间，我不能只使用User Role吗？ 我还尝试了clear-compile和composer dump-auto以及将Role引用替换为App\\Role并在顶部use App\\Role 。 我正在使用PHPStorm，它可以很好地引用参考，也可以从用户代码跳转到Role类定义。 在此先感谢您的帮助！

Answer 1

您无需传递类， Role是要在要扩展的类中调用的方法的字符串参数，因此您需要提供该类及其完整的名称空间。

return $this->belongsToMany('App\Role', 'users_roles');

Answer 2

你应该在做

return $this->belongsToMany('App\Role', 'users_roles');

和dump-auto和dump-autoload都相同。