Yii2：定义多个表之间关系的正确方法是什么？

Question

在控制器中我有以下代码：

public function actionView($id)
{
    $query = new Query;
    $query->select('*')
        ->from('table_1 t1')
        ->innerJoin('table_2 t2', 't2.t1_id = t1.id')
        ->innerJoin('table_3 t3', 't2.t3_id = t3.id')
        ->innerJoin('table_4 t4', 't3.t4_id = t4.id')
        ->andWhere('t1.id = ' . $id);
    $rows = $query->all();
    return $this->render('view', [
        'model' => $this->findModel($id),
        'rows' => $rows,
        ]);
}

查看数据库架构：https://github.com/AntoninSlejska/yii-test/blob/master/example/sql/example-schema.png

在视图 view.php 中显示来自 tables_2-4 的数据，这些数据与 table_1:

相关

foreach($rows as $row) {
    echo $row['t2_field_1'];
    echo $row['t2_field_2'];
    ...
}

参见：和：http://www.yiiframework.com/doc-2.0/yii-db-query.html

它有效，但我不确定它是否是最正确的 Yii2 方式。

我试图在模型 TableOne 中定义关系：

public function getTableTwoRecords()
{
    return $this->hasMany(TableTwo::className(), ['t1_id' => 'id']);
}
public function getTableThreeRecords()
{
    return $this->hasMany(TableThree::className(), ['id' => 't3_id'])
    ->via('tableTwoRecords');
}
public function getTableFourRecords()
{
    return $this->hasMany(TableFour::className(), ['id' => 't4_id'])
    ->via('tableThreeRecords');
}

然后在控制器TableOneController中加入记录：

$records = TableOne::find()
    ->innerJoinWith(['tableTwoRecords'])
    ->innerJoinWith(['tableThreeRecords'])
    ->innerJoinWith(['tableFourRecords'])
    ->all();

但它不起作用。如果我只加入前三个 table，那么它就可以工作。如果我添加第四个 table，则会收到以下错误消息："Getting unknown property: frontend\models\TableOne::t3_id"

如果我以这种方式更改函数 getTableFourRecords()：

public function getTableFourRecords()
{
    return $this->hasOne(TableThree::className(), ['t4_id' => 'id']);
}

然后我收到此错误消息：“SQLSTATE[42S22]：未找到列：1054 'on clause' 中的未知列 'table_4.t4_id' 正在执行的 SQL 是：SELECT table_1.* FROM table_1 INNER JOIN table_2 ON table_1.id = table_2.t1_id 内部联接 table_3 开启 table_2.t3_id = table_3.id 内部联接 table_4 开启 table_1.id = table_4.t4_id"

Answer 1

您应该必须在关系中定义键值对，例如：

class Customer extends ActiveRecord
{
    public function getOrders()
    {
        return $this->hasMany(Order::className(), ['customer_id' => 'id']); // Always KEY => VALUE pair this relation relate to hasMany relation
    }
}

class Order extends ActiveRecord
{
    public function getCustomer()
    {
        return $this->hasOne(Customer::className(), ['id' => 'customer_id']);
// Always KEY => VALUE pair this relation relate to hasOne relation
        }
    }

现在在你的第四个关系中使用：

public function getTableFourRecords()
{
    return $this->hasOne(TableThree::className(), ['id' => 't4_id']);
}

您可以在 ActiveRecord 上阅读更多内容 here

Answer 2

基于 softark 的answer，最简单的解决方案可能如下所示：

模型表一：

public function getTableTwoRecords()
    {
        return $this->hasMany(TableTwo::className(), ['t1_id' => 'id']);
    }

模型表二：

public function getTableThreeRecord()
    {
        return $this->hasOne(TableThree::className(), ['id' => 't3_id']);
    }

模型表三：

public function getTableFourRecord()
{
    return $this->hasOne(TableFour::className(), ['id' => 't4_id']);
}

控制器 TableOneController:

public function actionView($id)
{
    return $this->render('view', [
         'model' => $this->findModel($id),
    ]);
}

观点table-one/view.php:

foreach ($model->tableTwoRecords as $record) {
    echo ' Table 2 >> ';
    echo ' ID: ' . $record->id;
    echo ' T1 ID: ' . $record->t1_id;
    echo ' T3 ID: ' . $record->t3_id;
    echo ' Table 3 >> ';
    echo ' ID: ' . $record->tableThreeRecord->id;
    echo ' T4 ID: ' . $record->tableThreeRecord->t4_id;
    echo ' Table 4 >> ';
    echo ' ID: ' . $record->tableThreeRecord->tableFourRecord->id;
    echo ' <br>';
}

基于 GridView 的 解决方案也是可能的。

模型表二：

public function getTableOneRecord() { return $this->hasOne(TableOne::className(), ['id' => 't1_id']); } public function getTableThreeRecord() { return $this->hasOne(TableThree::className(), ['id' => 't3_id']); } public function getTableFourRecord() { return $this->hasOne(TableFour::className(), ['id' => 't4_id']) ->via('tableThreeRecord'); }

TableOneController 中使用 Gii 为模型 TableTwo 生成的函数 actionView 已编辑：

use app\models\TableTwo; use app\models\TableTwoSearch; ... public function actionView($id) { $searchModel = new TableTwoSearch([ 't1_id' => $id, // the data have to be filtered by the id of the displayed record ]); $dataProvider = $searchModel->search(Yii::$app->request->queryParams); return $this->render('view', [ 'model' => $this->findModel($id), 'searchModel' => $searchModel, 'dataProvider' => $dataProvider, ]); }

还有浏览量/table-one/view.php:

echo GridView::widget([ 'dataProvider' => $dataProvider, 'columns' => [ 'id', 't1_id', 'tableOneRecord.id', 't3_id', 'tableThreeRecord.id', 'tableThreeRecord.t4_id', 'tableFourRecord.id', ], ]);

参见code on Github。

Yii2：定义多个表之间关系的正确方法是什么？

Yii2: What is the correct way to define relationships among multiple tables?

inner-join

yii2