如何在 Yii2 gridview 中连接两个表并获取值 [英] How to join two tables and get values in Yii2 gridview

问题描述

我使用 Yii2 gridview widget 来显示数据.

I'm using Yii2 gridview widget to display the data.

我使用了两个名为 message 和 message_trigger 的表.

I'm using two tables named as message and message_trigger.

在message 表中的列是object_model、Object_id.

在message_trigger中的列是object_id，object_name.

网格从表格消息中获取值.所以网格字段是Object_model、Object_id.

The grid get the values from the table message. So grid fields are Object_model, Object_id.

现在我的问题是我需要根据表 message 中的 object_id 显示表 message_trigger 中的 Object_name强>.

Now my problem is I need to show Object_name from the table message_trigger based on the object_id from the table message.

在我的表单中，我使用了这样的网格

In my form I have used grid like this

<?= GridView::widget([
    'dataProvider' => $dataProvider,
    'filterModel' => $searchModel,
    'columns' => [
        ['class' => 'yiigridSerialColumn'],
        'object_model',
        'object_id',
        ['class' => 'yiigridActionColumn', 'template' => '{view} {update} {delete} '],
    ],
]); ?>

在我使用过的模型中

public function search($params){
    $query = AlertTrigger::find()->where(['alert_id'=>$params['id']])->andWhere(['!=','status',2]);
    $dataProvider = new ActiveDataProvider([
        'query' => $query,
    ]);
}

推荐答案

In Message Model

public function getMessageTrigger()
{
    return $this->hasOne(MessageTrigger::className(), ['object_id' => 'object_id']);
}

在视图中

<?= GridView::widget([
    'dataProvider' => $dataProvider,
    'filterModel' => $searchModel,
    'columns' => [
        ['class' => 'yiigridSerialColumn'],
        'object_model',
        'object_id',
        [
            'label' => 'Name',
            'value' => 'messageTrigger.object_name',
        ],
        ['class' => 'yiigridActionColumn', 'template' => '{view} {update} {delete} '],
    ],
]); ?>

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