Question

当我尝试使用ActiveDataProvider显示连接表（ hasOne 连接）时出现以下错误：

PHP Notice – yii\base\ErrorException
Trying to get property of non-object

我的模特：

public function dataProvider()
    {
        return new ActiveDataProvider([
            'query' => Currency::find()->with('country'),
            'pagination' => [
                'pageSize' => 20,
            ],
            'sort' => [
                'defaultOrder' => [
                    'allowed' => SORT_DESC,
                ]
            ],
        ]);
    }

hasOne 连接：

public function getCountry()
    {
        return $this->hasOne(Countries::className(), ['country_code' => 'country_code']);
    }

控制器：

public function actionManage()
    {
        return $this->render('manage', [
            'dataProvider' => Currency::find()->dataProvider(),
        ]);
    }

查看：

Pjax::begin(
    ['linkSelector' => 'a:not(.linksWithTarget)']
);

    echo GridView::widget([
        'dataProvider' => $dataProvider,
        'columns' => [
            [
                'label'  => 'Country',
                'value' => function ($model) {
                    return $model->country->country_name;
                },
            ],
        ],

    ]);

Pjax::end();

我在这里做错了什么？

Answer 1

亲爱的先生，您首先需要拥有关系名称国家/地区，如果您在数据库中设置关系，GII可以为您构建关系。如果不是你可以通过2找到简单地做到！（不推荐）

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<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery.scrollbar/0.2.11/jquery.scrollbar.js"></script>
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<h1>Width and height in style attr</h1>
<div class="scrollcontainer2 inner-scrollbar" style="width:200px;height:200px">
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</div>

Yii2 - ActiveDataProvider试图获取非对象的属性

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