不能在显示功能laravel中加入2个表格

Question

public function show($id){
 $users_services = DB::table('services')

->join('services','services.id', '=', 'users_services.service_id')
        ->get(array('users_services.id as id',
         'users_services.service_descreption as service_descreption',
        'users_services.service_link as service_link',
        'services.title as service_title','services.type as services_type'
        ))->toArray();
                                    ;   


    return  View('users_services.show')
    ->with('users_services', $users_services);

}

当我添加join语句时，出现此错误，并且我肯定有 users_services.service_id = services.id 我在视图中遇到此错误 “

  

试图获取非对象的属性“ services_title”（视图：C：\ new   xamp \ htdocs \ task \ digisay \ resources \ views \ users_services \ show.blade.php）

”

Answer 1

如下更改联接查询

   $users_services = DB::table('services')
     ->join('users_services','services.id', '=', 'users_services.service_id')
      ->select('users_services.id as id',
     'users_services.service_descreption as service_descreption','users_services.service_link as service_link',
    'services.title as service_title','services.type as services_type'
    )
    ->get();