Php 从模型控制器视图（laravel）传递联接查询

我试图通过DB连接进行循环，但出现错误：

Undefined variable: ContractorWorkTypePreferences

我在ContractorWorkTypePreference.php模型中创建了联接：

public static function workTypesSelected($contractorId){
   DB::table('contractor_work_type_preferences')
        ->join('work_types', 'contractor_work_type_preferences.work_type_id', '=', 'work_types.id')
        ->where('contractor_work_type_preferences.contractor_id', '=', $contractorId)
        ->groupBy('work_types.name')
        ->select('work_types.id', 'work_types.name')
        ->get();
} 

然后，我通过以下方式将其传递给Contractorcontroller.php：

public function show()
{

    //// get the contractor
    $arrPageData['contractor'] = Contractor::find($this->userID);   

    // get the work preference of the contractor
    $arrPageData['ContractorWorkTypePreferences'] = ContractorWorkTypePreference::workTypesSelected($this->userID);

    // show the create form and pass the contractor
    return View::make('contractors.show', $arrPageData);


}

最后，我尝试在视图show.blade.php中循环工作类型：

<ul class="tags-list">
            @foreach($ContractorWorkTypePreferences as $work)
            <li><a href="#">{{$work->name}}</a><li>
            @endforeach

        </ul>

@foreach（$ContractorWorkTypePreferences作为$work）

@endforeach

---

任何建议都将不胜感激。TIA.

看起来是打字错误<代码>承包商工作类型首选项和

承包商工作类型首选项

。错误到底发生在哪里？我试过单数和复数。错误发生在以下视图：@foreach（$ContractorWorkTypePreferences为$work）我也尝试了@foreach（$contractor->ContractorWorkTypePreferences为$work），这导致了错误：为foreach（）提供的参数无效首先，该静态方法上缺少

return

。错误是显而易见的，所以只需尝试将该变量重命名为不易出错的变量，然后重试。谢谢@deczo，我很感激。