Question

我创建了一个数据库助手来减少我在laravel框架中的代码，并连接到两个数据库，但是当我在助手中使用该函数时，我收到错误消息“试图获取非对象的属性”，这是我的助手代码：

/**
 * This function Used for get Data For Specific Element.
 */
public static function getDataById($dbName,$tableName,$condition,$data)
{
    $stattment=
        DB::connection($dbName)
            ->table($tableName)
            ->select(['*'])
            ->where($condition, $data)
            ->first();
    return $stattment;
}

这是我的控制器功能

public function all()
{

    $dataView['x']=dBHelper::allData('mysql',
        'products',
        '`status`=? AND `deleted`=?'
        ,array(1,1));

    if(is_object($dataView['x']))
    {
        foreach ($dataView['x'] as $key=>$value):
            $dataView['lang'][$key]=dBHelper::getDataById(
                'mysql2',
                'products',
                'id_product',
                $value->id);
            //var_dump($dataView['lang'][$key]);
        endforeach;
    }

    return view('productss.all',$dataView);
}

这是我的观点

<div class="content-wrapper">

    <!-- Content Header (Page header) -->

    <section class="content-header">

        <h1>

        </h1>

    </section>

    <!-- Main content -->

    <section class="content">

        <div class="row">

            <div class="col-lg-12">

                <div class="box box-primary">

                    <div class="box-header with-border">

                        <h3 class="box-title ">Show Products</h3>

                    </div>

                    <!-- /.box-header -->

                    <div class="adminform">



                        <!-- form start -->

                        <table id="all_data" class="table table_for_data table-striped table-bordered dt-responsive nowrap tableData" cellspacing="0" width="100%">

                            <thead>

                            <tr>
                                <th>Id</th>
                                <th>User Name</th>
                            </tr>
                            </thead>

                            <tbody>
                            @if(isset($x)&&is_object($x))
                                @foreach($x as $key=>$value)
                                    <tr>
                                    <td>{{$value->id}}</td>
                                    <td>{{$lang[$key]->name}}</td>
                                    </tr>
                                @endforeach

                            @else
                                <td>no data found</td>
                            @endif

                            </tbody>

                        </table>

                    </div>

                </div>

                <!-- /.box -->

            </div>

        </div>

    </section><!-- /.content -->

</div><!-- /.content-wrapper -->

我在打印时收到错误

<td>{{$lang[$key]->name}}</td>

谢谢。

Answer 1

将$dataView返回视图时，您应该：

@if(isset($dataView['x']) && is_object($dataView['x']))
    @foreach($dataView['x'] as $key=>$value)
        <tr>
            <td>{{$value->id}}</td>
            <td>{{$value['lang'][$key]->name}}</td>
        </tr>
    @endforeach

@else
    <td>no data found</td>
@endif

希望这会有所帮助。另一方面，像您这样的快捷方式使其难以调试或在将来的日期扩展。最好建立模型及其关系，从而利用Laravel的本机ORM。这会花费更多时间，但会大大加快应用程序的生产过程。

错误：尝试获取非对象的属性

1 个答案: