Php 无法在显示函数laravel中连接2个选项卡

}

当我添加join语句时，我得到了这个错误，我确信有 users\u services.service\u id=services.id 我在视图中看到了这个错误 "

正在尝试获取非对象的属性“services\u title”（视图：C:\new） xamp\htdocs\task\digisay\resources\views\users\u services\show.blade.php）

---

“

如下更改联接查询

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);

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转储结果数组并查看该数组中是否存在

services\u title

。可能您的对象没有

services\u title

属性。

   $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();