Yii 2：具有相同 table 的多重关系

Question

我遇到了 Yii 2 Relation Table 的问题。我的工作有很多关系，但只有在这种情况下return我一个错误：

SQLSTATE[42S22]: Column not found: 1054 Unknown column 'father.name' in 'where clause'

我认为问题是具有相同 Table "Agent" 的双重关系。见模型中的一段代码：

public function getAgent()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_agent']);
}
public function getFather()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_father']);
}

在我的 GridView 中，我看到了正确的值，但是当我尝试使用 ORDER 或 "andWhere" 进行过滤时，Yii2 return 出现了错误。

您可以在下面找到 searchModel 的代码片段：

$dataProvider->sort->attributes['agentName'] = [ 
        'asc' => ['agent.name' => SORT_ASC],
        'desc' => ['agent.name' => SORT_DESC],
        'default' => SORT_ASC
    ];

$dataProvider->sort->attributes['fatherName'] = [
        //'asc' => ['father.name' => SORT_ASC],
        //'desc' => ['father.name' => SORT_DESC],
        'default' => SORT_ASC
    ];
//.......
$query->andFilterWhere(['like', 'agent.name', $this->agentName]);
$query->andFilterWhere(['like', 'father.name', $this->fatherName]);

agentName 属性工作正常。有什么建议吗？谢谢！

--------更新：更多代码-------- 搜索模型：

public function search($params)
{
    $agent_aux = new Agent();
    $agent_id= $agent_aux->getAgentIdFromUser();

    if (Yii::$app->user->can('admin')){
        $query = Contract::find();
    }
    else{

        $query = Contract::find()->where(['contract.agent_id' => $agent_id]);            
    }

    $query->joinWith(['agent','seminar']);

    $dataProvider = new ActiveDataProvider([
        'query' => $query,
    ]);
    $this->load($params);
    $dataProvider->sort->attributes['seminar_location'] = [
        'asc' => ['seminar.location' => SORT_ASC],
        'desc' => ['seminar.location' => SORT_DESC],
    ];
    $dataProvider->sort->attributes['agentName'] = [ 
        'asc' => ['agent.name' => SORT_ASC],
        'desc' => ['agent.name' => SORT_DESC],
        'default' => SORT_ASC
    ];

    $dataProvider->sort->attributes['fatherName'] = [
        //'asc' => ['father.name' => SORT_ASC],
        //'desc' => ['father.name' => SORT_DESC],
        'default' => SORT_ASC
    ];
    if (!$this->validate()) {
        return $dataProvider;
    }
    $query->andFilterWhere([
        'id' => $this->id,
        'data' => $this->data,
        'id_agent' => $this->id_agent,
        'id_father' => $this->id_father,
        'id_seminar' => $this->id_seminar,
    ]);
    $query->andFilterWhere(['like', 'agent.name', $this->agentName]);
    $query->andFilterWhere(['like', 'father.name', $this->fatherName]);
    return $dataProvider;
}

Answer 1

您需要对模型进行以下更改。 from 子句实际上是在创建一个别名。代理人和父亲关系将在单独的连接子句中选择。在带有列名的过滤条件中使用 "agent" 和 "father" 别名。

public function getAgent()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_agent'])->from(['agent' => Agent::tableName()]);
}

public function getFather()
{
    return $this->hasOne(Agent::className(), ['id' => 'id_father'])->from(['father' => Agent::tableName()])
}

另一个要改变的是

$query->joinWith(['agent','seminar', 'father']);

Answer 2

@FidoXLNC 的答案的替代方法可能是在您进行连接时定义别名：

$query->joinWith([
    'seminar', 
    'agent'  => function ($q) { $q->from(Agent::tableName() . ' agent' ); },
    'father' => function ($q) { $q->from(Agent::tableName() . ' father'); }
]);

但是据我所知，您必须指定这两种关系，而不仅仅是一个。

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