我尝试使用infyom从关系对象数据库的表中使用scaffold，但我不能像POO中那样操作对象，因为生成的模型只包含其他对象的id！有没有办法正确操作物体？
生成的模型是这样的：

<?php

namespace App\Models;

use Eloquent as Model;

/**
 * Class Facture
 * @package App\Models
 * @version July 17, 2018, 6:08 pm UTC
 *
 * @property \Illuminate\Database\Eloquent\Collection Consomme
 * @property \Illuminate\Database\Eloquent\Collection contient
 * @property \Illuminate\Database\Eloquent\Collection EstFactureBst
 * @property \App\Models\EstFacture estFacture
 * @property string num_facture
 * @property date date_facture
 * @property string etat_facture
 * @property integer num_releve
 */
class personnel extends Model
{

    public $table = 'personnel';

    public $timestamps = false;
    public $sousPersonnel;

    protected $primaryKey = 'id_pers';

    public $fillable = [
        'id_pers',
        'lib_pers'
    ];

    /**
     * The attributes that should be casted to native types.
     *
     * @var array
     */
    protected $casts = [
        'lib_pers' => 'string'
    ];

    /**
     * Validation rules
     *
     * @var array
     */
    public static $rules = [

    ];

    public function SousPersonnel(){
        return $this->hasMany(\App\Models\TypePersonne::class);
    }
}

在我的类personnel和TypePErsonne之间有一个关系，但是当我想写像$personnel->typePersonne并使dd为null时，我使用Eloquent ORM，似乎他们无法为我的对象$personnel找到typePersonne